Assume as in the problem that each set $A$ or $B$ being considered is always sorted with elements indexed as $\{a_i\}_{i=1}^{100}$ and $\{b_i\}_{i=1}^{100}$, so if $i < j$, then $a_i < a_j$ and $b_i < b_j$. For each $i$ from $1$ to $100$, we can get a valid rule by comparing $a_i$ to $b_i$ and saying that $A$ wins if $a_i > b_i$ and $B$ wins if $a_i < b_i$. In other words, we're just taking the $i$th smallest element from every set and comparing sets by that element. It is easy to see each such rule satisfies the three conditions. We claim that these 100 rules are the only valid rules. Suppose we are given a rule. We will start by determining how the rule acts on pairs of sets $A, B$ where $a_i < b_j$ and $b_i < a_j$ whenever $i < j$. We'll call these pairs easy. Now, let $S = \{1, 2, \ldots, 100\}$ and define sets $T_i = \{t_{ij}\}_{j=1}^{100}$ where \[t_{ij} = \begin{cases} j - 0.1 &\mbox{if } j > i\\ j + 0.1 &\mbox{if } j \leq i \end{cases}, \] that is: \begin{align*} T_0 &= \{0.9, 1.9, 2.9, \ldots, 99.9\} \\ T_1 &= \{1.1, 1.9, 2.9, \ldots, 99.9\} \\ &\vdots \\ T_i &= \{1.1, 2.1, \ldots, i + 0.1, (i + 1) - 0.1, \ldots, 99.9\} \\ &\vdots \\ T_{100} &= \{1.1, 2.1, 3.1, \ldots, 100.1\}.\end{align*} By condition 2, $S$ beats $T_0$ but $T_{100}$ beats $S$. Therefore we can find $k$ so that $S$ beats $T_{k-1}$ but $T_k$ beats $S$. Now, let $A$ and $B$ be an easy pair of sets and suppose $a_k < b_k$. We prove that $B$ beats $A$: Let $\varepsilon$ be a positive real number smaller than $1/5$ the positive difference between any two members of $A \cup B$. Define $m_i = \min(a_i, b_i)$ and $M_i = \max(a_i, b_i)$, and construct the sets $A' = \{a'_i\}_{i=1}^{100}$ and $B' = \{b'_i\}_{i=1}^{100}$ where \[a'_{i} = \begin{cases} M_i + \varepsilon &\mbox{if } i < k\\ a_k + \varepsilon &\mbox{if } i = k\\ m_i - 2\varepsilon &\mbox{if } i > k \end{cases} \] and \[b'_{i} = \begin{cases} M_i + 2\varepsilon &\mbox{if } i < k\\ b_k - \varepsilon &\mbox{if } i = k\\ m_i - \varepsilon &\mbox{if } i > k \end{cases}. \] Then $B$ beats $B'$ because they are in the same relative order as $S$ and $T_{k-1}$, and $B'$ beats $A'$ by condition 2, and $A'$ beats $A$ because they are in the same relative order as $T_k$ and $S$. So by condition 3, $B$ beats $A$, as desired. This proves that if $A$ and $B$ are an easy pair of sets and $a_k < b_k$, then $B$ beats $A$. To generalize to arbitrary pairs of sets, you can, for example, interpolate sets $U_x = \{u_{xi}\}_{i=1}^{100}$ where $u_{xi} = (1-x)a_i + xb_i$ and change $x$ in sufficiently small increments.

The problem itself is related to this: https://en.wikipedia.org/wiki/Arrow%27s_impossibility_theorem

This proof is kind of lengthy (and clearly unnecessarily long) , but this is my idea and i hope that someone could verify whether is this correct or not. Thanks in advance! We will prove in general if the two players draw disjoint sets $A$ and $B$ of $n$ cards, there are exactly $n$ valid rules. Let $\mathcal{S}=\{S_1, S_2, \cdots, S_{M}\}$ be the set of all possible permutations of the relative positions of the $2n$ cards drawn by the two players, where $M=\dbinom{2n}{n}$. We define a rule as a function $f:\mathcal{S}\rightarrow\{\mathcal{A},\mathcal{B}\}$, in which $f(S_i)=\mathcal{A}$ if the permutation $S_i$ is $A$-win, or else $f(S_i)=\mathcal{B}$ if the permutation $S_i$ is $B$-win. Here we note the symmetry between the players, means that if permutation $S_j$ is obtained by exchanging all $a_i$'s in $S_i$ to $b_i$'s and all $b_i$'s in $S_i$ into $a_i$'s, then $f(S_i)\neq f(S_j)$. We note that for two given permutations of the relative positions of $A$ and $B$, and of $B$ and $C$, there can be several possible relative positions for $A$ and $C$, in which if $A$ beats $B$, $B$ beats $C$, then all these possible permutations must be $A$-win by Condition $1$ (We only know the relative position of $A$ to $B$, and $B$ to $C$, but not the exact values, so any possible permutation of $A$ to $C$ must have the same image under $f$ because we cannot tell what is the exact permutation of $A$ to $C$ without looking at the value of the cards). We would like to also define a set $\mathcal{T}=\{T_1, T_2, \cdots, T_{N}\}$, where $N=2^n$ and all elements in $\mathcal{T}$ are all binary strings with $n$ digits and consists of characters '$A$', '$B$' only. We assign a binary string in $\mathcal{T}$ to each permutations in $\mathcal{S}$ such that if the permutation $S_i$ satisfy $a_1>b_1$, then the first character of the binary string is '$A$', otherwise it is assigned with character '$B$', and if the permutation satisfy $a_2>b_2$, then the second character of the binary string is '$A$', otherwise it is assigned with character '$B$', and so on. Remark that two distinct permutations in $\mathcal{S}$ can be assigned with the same binary string in $\mathcal{T}$ (For example, both permutations $S_i=\{a_1,b_1,a_2,b_2,\cdots, a_n,b_n\}$ and $S_j=\{a_1,a_2,\cdots,a_n,b_1,b_2,\cdots b_n\}$ are distinct, but both assigned to the same binary string, namely '$AAA\cdots A$'). We will show that the binary string of a permutation will determine the image of the permutation under $f$. Lemma 1. If $S_i$ and $S_j$ are different by the position of $b_i$ between $a_i$ and $b_{i+1}$ only for one value of $i$, and both permutations have the same number of $a_i$'s between $a_i$ and $b_{i+1}$, then $f(S_i)=f(S_j)$. Proof. Let $S_k=\{ \cdots, a_i,b_i,a_{i+1},\cdots,a_j,b_{i+1},\cdots \}$, and suppose that the relative position of $A$ to $B$ is $S_k$, and the relative position of $B$ to $C$ is also $S_k$, then it suffices to show that $S_i$, $S_j$ are one of the possible relative positions of $A$ to $C$ (This will imply $f(S_i)=f(S_j)$ by Condition $3$). First we have the relative position of $A$ to $B$ is $\{\cdots, a_i,b_i,a_{i+1},\cdots,a_j,b_{i+1},\cdots\}$, and the relative position of $B$ to $C$ is $\{\cdots, b_i,c_i,b_{i+1},\cdots,b_j,c_{i+1},\cdots\}$. Since $a_{i+1},\cdots,a_j$ and $c_i$ is between $b_i$ and $b_{i+1}$, we know that the possible relative positions of $A$ to $C$ are $\{\cdots a_i, c_i, a_{i+1}, \cdots, a_j, c_{i+1}\cdots \}$, $\{\cdots a_i, a_{i+1}, c_i \cdots, a_j, c_{i+1}\cdots \}$, $\cdots$, $\{\cdots a_i,a_{i+1}\cdots, a_j, c_i,c_{i+1}\cdots \}$. Clearly, $S_i$ and $S_j$ must be one of these permutations, and each of these permutations must be all $A$-win or $C$-win by Condition $3$, depending on whether $S_k$ is $A$-win or $B$-win. So $S_i, S_j, S_k$ have the same image under $f$, which is $f(S_i)=f(S_k)=f(S_j)$, as required. Lemma 2. If two distinct permutations $S_i, S_j$ are assigned to the same binary string, then $f(S_i)=f(S_j)$. Proof. Let us consider the unique permutation $S_k$ which we call in the \emph{special} permutation for that binary string, such that for every $i$, if the $i$-th digit of the binary string is '$A$', then $a_i$ is at the $2i-1$-th position, and $2i$-th otherwise (For example, the binary string '$ABA$' gives the \emph{special} permutation $\{a_1,b_1,b_2,a_2,a_3,b_3\}$). We claim that for any permutation $S_i$ that is assigned to the same binary string, there exist a sequence of $n+1$ permutations $S_i,S_{i_1},S_{i_2},\cdots,S_{i_{n-1}},S_k$ in which all permutations are assigned to the same binary string, and satisfies $f(S_i)=f(S_{i_1})=f(S_{i_2})=\cdots=f(S_{i_{n-1}})=f(S_k)$. First, note that $S_i$ and $S_k$ must start with either $a_1$ or $b_1$, or else the first digit of the binary string of both permutations cannot be the same. WLOG let the first position of both permutations be $a_1$. Applying Lemma $1$ by choosing $a_1$ and $b_2$, we can 'move' $b_1$ to the $2$-nd position and we get a new permutation $S_{a_1}$, and this position of $b_1$ is clearly same as the position of $b_1$ in $S_k$. This new permutation $S_{a_1}$ clearly have the same binary string as $S_i$, and by Lemma $1$, $f(S_i)=f(S_{i_1})$. Now we look at the $3$-rd position, note that both permutations $S_{i_1}$ and $S_k$ must have either $a_2$ or $b_2$ at the $3$-rd position, and it has to be the same for both permutations, or else using similar reason as above, the second digit of the binary string of both permutations cannot be the same. Suppose the $3$-rd position for both permutations is $a_2$, then we do the same process by choosing $a_2$ and $b_3$ to 'move' $b_2$ to the fourth position to get the new permutation $S_{i_2}$. The case where the $3$-rd position is $b_2$ is completely analogous in which we move $a_2$ to the $4$-th position, and in both cases $f(S_{i_1})=f(S_{i_2})$ with same binary strings for both permutations. By repeating this process until $a_{n-1},b_{n-1}$, we can inductively find $S_{i_{n-1}}$ in which the first $2n-2$ positions are the same as $S_k$, and satisfy all conditions. However, similar to the argument for $a_1$ and $b_1$, if the $2n-1$-th position of $S_{i_{n-1}}$ and $S_k$ are not the same, then the last digit of the binary string of $S_{i_{n-1}}$ and $S_k$ are not the same, which implies the last digit of the binary string of $S_i$ and $S_k$ is not the same, false. So the $2n-1$-th and therefore the $2n$-th position of $S_{i_{n-1}}$ and $S_k$ must be the same, so $S_{i_{n-1}}=S_k$ and therefore $f(S_{i_{n-1}})=f(S_k)$. This means that for any $S_i$, we can find the \emph{special} permutation $S_k$ which is also assigned the same binary string and satisfies $f(S_i)=f(S_k)$. So for any two distinct permutations $S_i, S_j$ that are assigned to the same binary string, we have $f(S_i)=f(S_k)=f(S_j)$, as required. Therefore, Lemma $2$ implies that we can now define a new function $g:\mathcal{T}\rightarrow \{\mathcal{A},\mathcal{B}\}$ in which the function $f$ completely depends on $g$, and $g$ completely depends on the binary strings in $\mathcal{T}$ of all the permutations in $\mathcal{S}$ . Due to symmetry again, given two distinct binary strings $T_x,T_y \in\mathcal{T}$, if the first digit of $T_x$ and $T_y$ is different, second digit of $T_x$ and $T_y$ is different, and so forth until the $n$-th digit of $T_x$ and $T_y$ is different, then $g(T_x)\neq g(T_y)$ . This means that exactly $2^{n-1}$ binary strings in $\mathcal{T}$ must have $\mathcal{A}$ be the image under $g$, so as to $\mathcal{B}$. Let us denote a binary string $T$ to be the \emph{A-friend} of $T_1, T_2, \cdots T_k$ such that the $i$-th digit of $T$' is '$A$' if and only if the $i$-th digit of all $T_1, T_2, \cdots T_k$ are '$A$' (For example, the \emph{A-friend} of '$ABABA$', '$ABBAB$' and '$ABBAA$' is '$ABBBB$'). Note that we can similarly define \emph{B-friend} due to symmetry, which the 'roles' of '$A$' and '$B$' is swapped, but we will not use it in this solution. Lemma 3. Given binary numbers $T_1, T_2,\cdots , T_k$ that satisfy $g(T_1)=g(T_2)=\cdots =g(T_k)$. Let $T$ be the \emph{A-friend} of the binary strings $T_1, T_2, \cdots, T_k$, then $g(T)=g(T_1)=g(T_2)=\cdots =g(T_k)$. Proof. We first solve the case where $k=2$, where $T_z$ is the \emph{A-friend} of binary strings $T_x, T_y$. We again look at the \emph{special} permutation of the binary string. We can divide the \emph{special} permutations into $n$ pairs (first and second, third and fourth, and so on), each consisting of $a_i$ and $ b_i$ for every $1\le i\le n$ (This is by the definition of a \emph{special} permutation that the $n$ pairs we divide will not contain both $a_i$'s or both $b_i$'s). Similar to the proof of Lemma $1$, we let the relative position of $A$ to $B$ be the \emph{special} permutation of $T_x$, and let the relative position of $b$ to $c$ be the \emph{special} permutation of $T_y$. Now for every $i$, if the $i$-th digit of $T_x, T_y$ are both '$A$', then this digit correspond to $a_i>b_i$ and $b_i>c_i$, and thus $a_i>c_i$, corresponds to the '$A$' in $T_z$. The case where the $i$-th digit of $T_x, T_y$ are both '$B$ is analogous. However, if the $i$-th digit of $T_x, T_y$ are different, then this corresponds to either $a_i>b_i$, $c_i>b_i$ or $a_i<b_i$, $c_i<b_i$, and in both cases the possible relative positions for $a_i$ to $c_i$ are $a_i>c_i$ and $a_i<c_i$. We choose the case where $a_i<c_i$ for every such $i$ (where the $i$-th digit of $T_x, T_y$ are different), and these correspond to the digits '$B$' in $T_z$. By Condition $3$ again, this permutation must have same image with the \emph{special} permutation of $T_x$ and $T_y$ under $f$, so $T_z$ must have the same image with $T_x$ and $T_y$ under $g$, that is $g(T_z)=g(T_x)=g(T_y)$. Now we solve for the general case. It is obvious that if $T$ is the \emph{A-friend} of $T_1, T_2, \cdots, T_k$ and $T'$ is the \emph{A-friend} of $T_{k+1}$ and $T$, then $T'$ is also the \emph{A-friend} of $T_1, T_2, \cdots, T_{k+1}$. So if $T_{i_1}$ is the \emph{A-friend} of $T_1$ and $T_2$, then from the above paragraph $g(T_{i_1})=g(T_1)=g(T_2)$ must hold. Let $T_{i_2}$ be the \emph{A-friend} of $T_{i_1}$ and $T_3$, then $g(T_{i_2})=g(T_3)=g(T_2)=g(T_1)$. Repeat this until we get $T_{i_{k-1}}$ be the \emph{A-friend} of $T_{i_{k-2}}$ and $T_k$, and since the \emph{A-friend} of $T_{i_{k-2}}$ and $T_k$ is actually the \emph{A-friend} of $T_{i_{k-3}}$, $T_{k-1}$ and $T_k$, and so on, until $T_{i_{k-1}}$ is the \emph{A-friend} of $T_k, T_{k-1},\cdots T_1$, which is $T$. So $T=T_{i_{n-1}}$ and thus we get $g(T)=g(T_{i_k})=g(T_k)=\cdots =g(T_2)=g(T_1)$, as required. Lemma 4. Let $T_1, T_2, \cdots T_{2^{n-1}} \in \mathcal{T}$ be the set of $2^{n-1}$ binary strings which have image $A$ under $g$. Then there must be exactly one common digit among all $2^{n-1}$ binary strings, and this common digit must be '$A$'. Proof. Suppose the contrary holds. We first note that from Condition $2$, $a_i>b_i$ for all $i$ is $A$-win, which correspond to the binary string '$AAA\cdots AA$', so $g($'$AAA\cdots AA$'$)=\mathcal{A}$. Due to symmetry again, we have $g($'$BBB\cdots BB$'$)=\mathcal{B}$. We consider the \emph{A-friend} $T$ of $T_1, T_2, \cdots T_{2^{n-1}}$. Then from the hypothesis $T=$'$BBB\cdots BB$', or else suppose the $i$-th digit of $T$ is '$A$', then the $i$-th digit of all the $T_1, T_2, \cdots T_{2^{n-1}}$ is '$A$', which contradicts the assumption that there are no common digit among $T_1, T_2, \cdots T_{2^{n-1}}$. Then since $g($'$AAA\cdots AA$'$)=\mathcal{A}$, the binary string '$AAA\cdots AA$' must be one of the binary strings $T_1, T_2, \cdots T_{2^{n-1}}$, then by Lemma $3$, $\mathcal{A}=g($'$AAA\cdots AA$'$)=g($'$AAA\cdots AA$'$)=\mathcal{B}$, clearly a contradiction. If the common digit is '$B$', then the fact that the binary string '$AAA\cdots AA$' is one of the binary strings $T_1, T_2, \cdots T_{2^{n-1}}$ gives a contradiction, and also by Pigeonhole Principle, it is impossible to have at least $2$ common digits since there are only $2^{n-k}$ binary strings that have $k$ common digits. It is therefore clear that if $g$ is a valid function, then there exist exactly one common digit among all $2^{n-1}$ binary strings which have image $A$ under $g$, and this digit must be '$A$', as required. Therefore by Lemma $4$, $g$ is completely dependent on the choice of a unique positive integer in the set $\mathcal{U}=\{1,2,\cdots, n\}$, in which if a unique positive integer $j \in \mathcal{U}$ is chosen, then we get the function $g_j: \mathcal{T}\rightarrow \{A,B\}$ that satisfy the condition that if the $j$-th digit of a binary string $T_x$ is '$A$', then $g(T_x)=\mathcal{A}$. Since the set $\mathcal{U}$ has $n$ elements, the number of valid functions $g$ is at most $n$. Note that by the original definition of a binary string, then for every $j$, the function $g_j$ that maps every binary string with $j$-th digit is '$A$' to its image $\mathcal{A}$ actually corresponds to the rule 'A beats B iff $a_j>b_j$ for that $j$'. It remains to verify these $n$ rules are valid rules. These rule satisfy Condition $1$ because it depends only on the relative position of the $2n$ values drawn by the players, and it also satisfy Condition $2$ since if $a_i>b_i$ for all $i$, then it clearly implies $a_j>b_j$ for that $j$. Finally, since $A$ beats $B$, $B$ beats $C$ if and only if $a_j>b_j$ and $b_j>c_j$ , then $a_i>c_i$ which implies $A$ beats $C$, so it satisfies Condition $3$. We conclude that there are exactly $n$ valid rules for every $n$. Set $n=100$, then the desired answer is $100$. Q.E.D The idea is essentially a double mapping that maps all the permutations to binary strings with length $n$, and maps to a positive integer from $1$ to $n$.

Say that a set $X$ is ranked if $X=\{x_1,x_2,\dots ,x_{100}\}$ is a set of integers with $1\le x_1< x_2\dots , x_{100}\le 200$, and define $\overline{X}=\{1,2,\dots , 200\}\setminus X=\{\overline{x_1},\overline{x_2},\dots, \overline{x_{100}}\}$ with $1\le \overline{x_1}<\overline{x_2}\dots < \overline{x_{100}}\le 200$ to be the complement of the set $X$. Define $X$ to be winning if the set $X$ wins against $\overline{X}$, and define $X$ to be losing otherwise. Also, for a ranked set $X$, define its representative as the $100$-tuple $(r_1,r_2,\dots, r_{100})$, where $r_i=0$ if $x_i>y_i$ and $r_i=-1$ otherwise. Finally, say that for a representative that its linchpin is the ranked set $\{2i+r_i: 1\le i\le 100\}$, which clearly has magnitude $100$, and that this linchpin has a plus set $P=\{i:r_i=0\}$. Furthermore, let $A$ and $B$ be disjoint sets of distinct real numbers $A=\{a_1, a_2, \dots , a_{100}\}$ with $a_1<a_2<\dots <a_{100}$ and $B=\{b_1, b_2, \dots , b_{100}\}$ with $b_1<b_2<\dots <b_{100}$, and let $A\cup B=S=\{s_1, s_2, \dots , s_{200}\}$ with $s_1<s_2< \dots <s_{200}$. Define the normalized set $N(A)$ with respect to $B$ to be $\{j:1\le j\le 200 \text{ and } s_j=a_i \text{ for some } 1\le i\le 100\}$. By condition $1$, it follows that $N(A)$ with respect to $B$ is winning if and only if $A$ wins over $B$. For a set $S$, a real number $c$, and an integer $k$, define $S+c=\{s+c:s\in S\}$ and the subset $S_k$ of $S$ to consist of exactly the elements of $S$ that are at least $k$. Lemma 1. $X$ and $Y$ are two ranked sets $x_i\ge y_i$ over all $1\le i\le 100$. If $Y$ is a winning set, then so is $X$. Proof 1. By condition $2$, since $x_i+.1>y_i$, it follows that $X+.1$ wins over $Y$. Since $Y$ is a winning set, $Y$ wins over $\overline{Y}$. Note that $x_i\ge y_i$ for all $i$ implies that $|X_k|\ge |Y_k|$ over all integers $k$. But it then follows that $|\left(\overline{X}\right)_k|\le |\left(\overline{Y}\right)_k|$ over all integers $k$, so hence $\overline{x_i}\le \overline{y_i}$ for all $1\le i\le 100$. Then, $\overline{y_i}>\overline{x_i}-.1$, so $\overline{Y}$ wins over $\overline{X}-.1$ by condition $2$. Hence, by condition $3$, it follows that $X+.1$ wins over $\overline{X}-.1$, but since the elements of $X\cup \overline{X}=\{1,2,\dots , 200\}$ all differ from one another by at least $1$, it follows from condition $1$ that $X$ wins over $\overline{X}$, so hence $X$ is a winning set. $\square$ Lemma 2. Let $X$ be a ranked set, and consider its representative $R$. We define another ranked set $Y$ as follows: If $r_i=0$, then $y_i=2i$. Otherwise, if $r_i=-1$, let $j$ be the largest integer less than $i$ for which $r_j=0$ (Here, define $r_0=0$), then $y_i=i+j$. Call $Y$ the dominated set for the representative $R$. If $Y$ is winning, then so is $X$. Proof. If $r_i=0$, then $x_i>x_1, x_2, \dots x_{i-1}, \overline{x_1}, \overline{x_2}, \dots , \overline{x_i}$, so $x_i\ge 2i$. If $r_i=-1$, then $x_i>x_i, x_2, \dots x_{i-1}, \overline{x_1}, \overline{x_2}, \dots , \overline{x_j}$, so $x_i\ge i+j$. By Lemma $1$, it follows that if $Y$ is winning, then so is $X$. $\square$ Let $L$ be a linchpin. Define $L=\{l_{0,1}, l_{0,2}, \dots , l_{0,100}\}$ and $L_1=\overline{L}$, and we will define the sets $L_k=\{l_{k,1}, l_{k,2}, \ldots , l_{k,100}\}$ for integers $2\le k\le 100$ in such a way that $N(L_i)$ with respect to $L_{i+1}$ is simply $L$ for all $i\ge 0$. Lemma 3. Let $1\le i\le 99$ be an integer such that the representative of $L$ has $r_i\neq r_{i+1}$. Then, there exists a constant $c$ satisfying $2i<c<2i+1$ for which $l_{k,i}<c<l_{k,i+1}$ for all positive integers $k$. Proof. When $r_i=-1$ and $r_{i+1}=0$, because $N(L_i)$ with respect to $L_{i+1}$ is simply $L$ for all $i\ge 0$, it follows that $l_{0,i} < l_{1,i}=2i < \dots < l_{100,i} < l_{100,i+1} < \dots <l_{1,i+1}=2i+1 < l_{0,i+1}$. When $r_i=0$ and $r_{i+1}=-1$, it follows that $l_{100,i}<\dots <l_{1,i}<l_{0,i}=2i$ and $l_{0,i+1}=2i+1<l_{1,i+1}<\dots <l_{100,i+1}$, so any choice of $2i<c<2i+1$ will work. $\square$ Lemma 4. Let $X$ be a ranked set with the same representative as $L$. Then, $X$ is winning if and only if $L$ is winning. Proof. Let $\epsilon <\dfrac{1}{200}$ be a small positive number. Break up the representative of $L$ into blocks of consecutive $0$'s and $-1$'s. By Lemma $3$, we may focus on a single block. Consider any block of $-1$'s. For simplicity's sake, consider it to be $l_1, l_2, \dots , l_b$. For each $i=1,2, \dots b$ and $k\ge 2$, define $l_{k,i}$ as follows: if $k\le b+1-i$, $l_{k,i}=l_{k-1,i}+2-\epsilon$; otherwise, $l_{k,i}=l_{k-1,i}+\epsilon$. It can be checked that for all such $k$, it holds that $l_{k-1,1}<l_{k,1}<l_{k-1,2}<l_{k,2}<\dots <l_{k-1,b}<l_{k,b}$. $(*)$ Then, $l_{100,1}=2+(b-1)(2-\epsilon)+(100-b)\epsilon>2b-1=l_{0,b}$. On the other hand, for any $r_i=0$, simply define $l_{k,i}=l_{k-1,i}-\epsilon$. This construction makes it true that $N(L_{k-1})$ with respect to $L_k$ is in fact $L$, and it can be seen that $\epsilon$ is chosen such that even when $r_j=-1$ and $r_{j+1}=0$, still $l_{100,j}<l_{100,j+1}$. Assume that $L$ is winning, with representative $R$. Then, by condition $3$, it follows that $L$ wins over $L_{100}$. By using $(*)$, it can be seen that $N(L)$ with respect to $L_{100}$ is just the dominated set of $R$. Thus, since the dominated set of $R$ is winning by condition $1$, it follows that $X$ itself must be winning by Lemma $2$. On the other hand, if $L$ is losing, then $\overline{L}$ (which is the linchpin of the representative of $\overline{X}$) is winning, so $\overline{X}$ is winning, and $X$ is losing. Hence, the lemma is proved. $\square$ Now, let $L=(l_1, l_2, \cdots l_{100})$ and $M=(m_1, m_2, \cdots m_{100})$ be two distinct linchpins that are winning, and let $P$ and $Q$ be their respective plus sets. Lemma 5. The linchpin whose plus set is $P\cap Q$ is also winning. Proof. Let $\epsilon '<\dfrac 12$ be a small positive number. We will define a set $S=\{s_1, s_2, \dots , s_{100}\}$ with $s_1<s_2<\dots s_{100}$, so that $s_i$ is based on cases whether $i\in P$ and $i\in Q$. If $i\in P$ and $i\notin Q$, then let $s_i=\overline{p_i}+(1+\epsilon ')$; if $i\notin P$ and $i\notin Q$, then let $s_i=\overline{p_i}+\epsilon '$; and if $i\in Q$, then let $s_i=\overline{p_i}-\epsilon '$. It then follows that $N(\overline{P})$ with respect to $S$ is $Q$, and by condition $3$, $P$ wins over $S$. It can be seen that $N(P)$ with respect to $S$ is a winning set by condition $1$, and is the linchpin whose plus set is $P\cap Q$, so the lemma is proven. $\square$ Hence, if $T$ is the intersection of the plus sets over all winning linchpins, it follows that only those linchpins whose plus set is a superset of $T$ can be winning. Furthermore, by Lemma 1, it follows that all of these linchpins are in fact winning; hence the number of winning linchpins is $2^{100-|T|}$. However, since the linchpins can be paired up into $2^{99}$ complementary ranked sets, there are a total of $2^{99}$ winning linchpins. Hence, $|T|=1$, so there are at most $100$ possible rules for this game because each choice of $T$ uniquely determines whether each linchpin is winning, which in turn determines whether each ranked set is winning, which completely determines the game by condition $1$. We can easily construct these rules so that all of the conditions are satisfied; that is, for each integer $1\le i\le 100$, we let the $i$th rule be to let the set $A$ to win over $B$ if and only if $a_i>b_i$. These $100$ possible rules satisfy all the conditions, so the answer is $100$.

There are $100$ ways to define such a rule. For each $i, 1 \le i \le 100$, define a rule as follow: let $A_i$ and $B_i$ be the $i$ th biggest element of $A$ and $B$, then $A$ wins iff $A_i > B_i$. It is easy to see that any of these work. Now we take any rule that satisifies the problem's condition and show that it has to be one of the $100$ rules we just stated. For two sets $A$ and $B$, say $A > B$ if $A$ is the winner when $A$ and $B$ are compared. For a set $X$ of $100$ numbers $X_1, \cdots X_{100}$ (sorted in ascending order), we consider some set $Y$ with $Y<X$ and with the biggest amount of number of elements, $s$, bigger than $X_{100}$.

We assume $s >0$ (the case $s = 0$ is easy to see). We have something like this: \[ \overbrace{\quad X_1\quad X_2 \quad \cdots }^\text{contains smallest 100-s elements of Y} X_{100} \quad Y_{101-s}\quad \cdots \quad Y_{100} \quad \quad (Y<X)

\[ \Rightarrow \label{blah} Y_1\quad Y_2 \quad \cdots \quad Y_{100-s} \quad Z_1 \overbrace{ \quad Z_2 \quad \cdots Z_{100} \quad}^\text{contains biggest s elements of $Y$}\quad \quad (Y<Z) \qquad \qquad \qquad \qquad (1)\]2. Claim: If a set $Y$ loses to a set $X$ and has $s$ biggest elements to the right of biggest element of $X$, then the $s+1$th biggest element must be before the $s+1$th biggest element of $X$.

Thus by contrapositive of the claim: \[X_1 \quad X_2 \quad \cdots \quad X_{100-s} \quad \cdots \quad Y_{100-s} \overbrace{\cdots \quad X_{99} \quad X_{100}}^\text{at most s elements from X} Y_{101-s} \quad \cdots \quad Y_{100}\quad \quad \Rightarrow (X<Y) \qquad \qquad \qquad \qquad (2)\] After substituting $Z$ into $Y$ and $X$ into $Z$ in $(1)$ and comparing $(1)$ and $(2)$ we see that $Z<Y$ as long as the $100-s$th smallest element of $Z$ is smaller than $100-s$th smallest element of $Y$.

Quick solution sketch: Notice that the transitive property is basically comparing the 3 relations between A,B and B,C and A,C. If we make the relation between A,B and B,C to be the same, then the relation A,C must have the same outcome. Then we search for an invariant and find: Write out the 200 numbers in decreasing order and pair up $a_i$ and $b_i$ and draw the arrow pointing from $a_i$ to $b_i$. One can assign a n digit binary string to this arrangement by having the ith digit be 1 if the ith arrow is pointing right, 0 otherwise. Using the method from before, we get that invariant, so we should conjecture that all arrangements with same binary string are same. Then we know that 00...0 is losing, 11...1 is winning. Also we can just consider the binary string and not the cards.(if works for 1 arrangement with binary string, works for all arrangements with binary string) We can construct A,B,C s.t. the binary string corresponding to A,C relation is the or function of the binary strings of the relations A,B and B,C.(This is nice, since now we just have binary strings, the or function, and no cards! So is ez counting from here) Thus, the rule works only if some index is 1 for all winning arrangements. So we have 100 winning arrangements. Note: How does somebody do this within 2 hours in contest?

Thats the only point of IMO.

We claim that the answer is $\boxed{100}$. Let the $k$-rule be the rule that says $A$ wins if and only if $a_k>b_k$ where the sets are ordered as given in the problem. It is trivial to verify that the $k$-rule is valid, so it remains to show that there are no other rules. Indeed, suppose for sake of contradiction, suppose there was a rule that is not the $k$-rule for any value of $k$. Lemma: Assume \[a_1>\cdots>a_{k-1}>b_1>\cdots>b_k>a_k>\cdots>a_{100}>b_{k+1}>\cdots>b_{100}.\]Then $A$ wins. Proof: Suppose to the contrary that $B$ wins. Suppose we place $c_1>\ldots>c_{100}$ such that $a_i>c_i$ for all $i$. It is easy to see that we may get any relative ordering of $B$ and $C$ subject to the condition $b_k>c_k$, and the third condition of the problem implies that $B$ beats $C$, so the $k$-rule is valid, which is the desired contradiction. $\blacksquare$ This allows us to prove the following inductive claim which clearly gives a contradiction. We refer to the $k$th instance of the lemma as $L_k$. Claim: Fix any $1\le\ell\le 100$. If $A$ and $B$ are arranged such that $b_{\ell+1}<a_{100}$, then $A$ wins. Proof: We induct on $\ell$ with $\ell=1$ being $L_1$. Now suppose the claim is true for $\ell$. Apply $L_{\ell+1}$ to $B$ to generate a $C$ such that $B$ beats $C$ (according to $L_{\ell+1}$ with $A\to B$ and $B\to C$). It is easy to see that we can get any relative ordering of $A$ and $C$ subject to $c_{\ell+2}<a_{100}$, so we're done by the third condition. $\blacksquare$ Note that the $\ell=100$ instance of the claim is clearly a contradiction, so we're done.

$\boxed{\text{Answer =} \; 100}$. Solved with a senior from IMO 2015-2016. Proof by Contradiction mixed up with circular optimization: 100 contradictions against one truth. Call the sets of 100 cards (that the players take) to be $\textit{decks}$ for writeup ease. In this proof, I will frequently say that the card with number $1$ is bigger than the card with number $2$, and the deck $(1,3)$ will beat $(2,4)$. I initially thought of the cards in a rank system, with the leftmost cards in the number line taking priority, and I didn't realize that the problem implicitly meant the $\textit{actual}$ real line --- so there's that. $\color{green} \rule{25cm}{1pt}$ $\color{green} \textbf{First Part.}$ The 100 rules are define-able easily: in the $i^{th}$ rule, one position beats the other if and only if its $i^{th}$ card, sorted by size, is bigger than the other position's $i^{th}$ card. It's easy to see that this rule is clearly defined and satisfies both three conditions. $\blacksquare$ $\color{red} \rule{25cm}{2.5pt}$ $\color{red} \textbf{Second Part.}$ We will prove that all other rules yield contradictions --- let there exists another rule which isn't equal to the 100 rules we established. Then, in that rule, for each $1 \leq i \leq 100$, there is a pair of decks $P_i, Q_i$ where \[\begin{tabular}{p{12cm}} $P_i$ beats $Q_i$, with $P_i$'s $i^{th}$ card smaller than $Q_i$'s $i^{th}$ card. \\ \end{tabular} \\ \]Let $R_i$ to be a much weaker deck than $Q_i$, that is the deck formed by stacking all cards bigger than the $i^{th}$ card near it by sufficiently small $\epsilon$ (a.k.a. $\epsilon$ smaller than the difference of the $i^{th}$ card and the smallest card bigger than it) and stacking all cards smaller than the $i^{th}$ card below the minimum of both decks. Since by condition (2) $Q_i$ beats $R_i$, so $P_i$ beats $R_i$. This is equivalent to saying that the deck with cards in ranks $(1,2,3,\ldots,i-1,2i,2i+1,\ldots,100+i)$ beats the deck $(i,i+1,\ldots,2i-1,101+i,\ldots,200)$. We will prove that we can work with these 100 almost normal anomalies to establish a contradiction. Here we milk condition (3) to its fullest, and land the final blow using the condition (2) against itself. First let a dummy deck $D_0=(1,2,3,\ldots,100)$. Fix $\frac{100(100+1)}{2}$ numbers between $99$ and $100$, and do the same between $0$ and $1$. Out of these $100 \cdot 101$ numbers we will create $101$ decks $D_1,D_2,\ldots,D_{101}$ so that $D_i$ beats $D_{i+1}$ for every $0 \leq i \leq 101$ (taken mod $102$), effectively forming a contradiction. Construct $D_i$ step-by-step; if $D_1,D_2,\ldots,D_{i-1}$ is already constructed, $D_{i}$ will take the leftmost $i-1$ numbers not already taken from the interval $(0,1)$ and the leftmost $n+1-i$ numbers not already taken from the interval $(99,100)$. Now, the relations \[ D_0 \rightarrow D_1, D_1 \rightarrow D_2, \ldots, D_{99} \rightarrow D_{100}, D_{100} \rightarrow D_{101} \]is established due to the aforementioned relations between $P_i$ and $R_i$: \[ P_{100} \rightarrow R_{100}, P_1 \rightarrow R_1, \ldots, P_{99} \rightarrow R_{99}, P_{100} \rightarrow R_{100} \]The ultimate victory between $D_{101}$ against $D_0$ is established by noticing that all of $D_{101}$'s numbers are greater than the greatest of $D_0$'s numbers. We are done. $\blacksquare$ $\blacksquare$ $\blacksquare$

Solved with Alan Bu, Chris Qiu, Edward Yu, Elliott Liu, Eric Shen, Isaac Zhu, Kevin Wu, Marvin Mao, Ryan Yang The answer is 0. Assume for the sake of contradiction that there exists a rule. Then, there must exist a deck of cards; which must be countable. However, the set of real numbers is uncountable, so such a deck can not exist. We conclude that no rules exist.

Solved with Alan Bu, Chris Qiu, Edward Yu, Elliott Liu, Isaac Zhu, Jeffrey Chen, Kevin Wu, Marvin Mao, and Ryan Yang. The answer is \(n=100\), achieved by each rule ``\(A\) beats \(B\) if and only if \(a_i>b_i\)'' for \(1\le i\le n\). Now we prove these are the only rules that work. Say \(A\prec B\) whenever \(B\) beats \(A\). It suffices to find an \(i\) with the following property: For \(A=\{a_1<a_2<\cdots<a_n\}\) and \(B=\{b_1<b_2<\cdots<b_n\}\), we have \(A\prec B\) if and only if \(a_i<b_i\). I claim we can take \(1\le i\le n\) minimal satisfying \[(1,\ldots,i,n+i+1,\ldots,2n)\prec(i+1,\ldots,i+n).\]Such \(i\) exists, since when \(i=0\), we have \((n+1,\ldots,2n)\succ(1,\ldots,n)\), and when \(i=n\), we have \((1,\ldots,n)\prec(n+1,\ldots,2n)\). For \(A=\{a_1<a_2<\cdots<a_n\}\) and \(B=\{b_1<b_2<\cdots<b_n\}\) with \(a_i<b_i\), construct \begin{align*} A'&=\{a_1+\varepsilon<\cdots<a_{i-1}+\varepsilon<a_i+\varepsilon<\infty<\cdots<\infty\}\\ B'&=\{-\infty<\cdots<-\infty<b_i-\varepsilon<b_{i+1}-\varepsilon<\cdots<b_n-\varepsilon\}. \end{align*}Note that \(A\prec A'\) and \(B'\prec B\) by design. Finally, take \(X=\{x_1<\cdots<x_n\}\) satisfying \[a_i+\varepsilon<x_1<\cdots<x_n<b_i-\varepsilon.\]Observe that: Since \(\{1,\ldots,i,n+i+1,\ldots,2n\}\prec\{i+1,\ldots,i+n\}\), it follows that \(A'\prec X\). Since \(\{1,\ldots,i-1,n+i,\ldots,2n\}\succ\{i,\ldots,i-1+n\}\) by minimality of \(i\), it follows that \(X\prec B'\). It follows that \(A\prec A'\prec X\prec B'\prec B\), and we are done.

The answer is 100. The only rules are as follows: fix $1\le k\le 100$. A beats B if and only if $a_k>b_k$. It's clear that they work; now I show they are the only solutions. Replace beat with majorize. Say $A=\{a_1, \cdots, a_{100}\} $ majorizes $B$. Also WLOG the range of the cards is reals (also a countable set won't change the main structure of the argument) Claim: [We can get a lot of information fron juggling] suppose $b_{t_0}<\cdots < b_{t_1-1} \\ <a_1 < b_{t_1} < \cdots <b_{t_2-1} < \cdots < b_{t_i-1} < a_i < b_{t_i} < \cdots < b_{t_n-1}< a_n< b_{t_n} < \cdots $ then $B$ majorizes $A$ but by swapping $a_i$ with $b_{t_i}$, $A$ majorizes $B$, then $a_1<b_1<\cdots<b_{i-1}<b_i<a_i<a_{i+1}<\cdots<b_n$. Proof: suppose $b_{t_j-1} < a_j-\epsilon < a_j< b_{t_j}$ and $b_{t_i-1}<a_i-\epsilon< b_{t_i} < a_i< b_{t_i+1}$ for all notice $(a_1-\epsilon, \cdots, a_{i-1}-\epsilon, a_i, a_{i+1}-\epsilon, a_n-\epsilon)$ majorizes $(b_1, \cdots, b_n)$, which in turn majorizes $(a_1, \cdots, a_{i-1}, a_i-\epsilon, a_{i+1}, \cdots, a_n)$ Therefore, if $a_1<b_1<\cdots<b_{i-1}<b_i<a_i<a_{i+1}<\cdots<b_n$ then $A$ majorizes $B$. By symmetry, it also follows if $a_1<b_1<\cdots<b_{t_i-1}<b_{t_i}<a_{t_i}<a_{t_i+1}<\cdots<b_n$ then $A$ majorizes $B$. (End of (M)) Lemma 2: if $i\ne j$, $a_1<b_1<\cdots<b_{i-1}<b_{i}<a_{i}<a_{i+1}<\cdots<b_n$ implying $A$ majorize $B$ and $a_1<b_1<\cdots<b_{j-1}<b_{j}<a_{j}<a_{j+1}<\cdots<b_n$ implying $A$ majorize $B$ cannot simultaneously occur. Proof: $(a_1,a_2,\cdots,a_n)$ majorizes $(a_1+\epsilon, a_2+\epsilon, \cdots, a_i-\epsilon/2, a_{i+1}+\epsilon, \cdots, a_n+\epsilon)$ which majorizes $(a_1+2\epsilon, \cdots, a_i+\epsilon/2, \cdots, a_{t_i}+\epsilon/2, \cdots, a_n+2\epsilon)$ which contradicts property 2. By lemma 2, $t_i=i$. Consider the following process: Start with $a_1<b_1<\cdots<b_{i-1}<b_i<a_i<\cdots<b_n$. Swap two adjacent elements of two different sequences, say $b_y,a_x$ are adjacent terms in the sorted sequence of $A\cup B$ and I give $b_y$ to $A$ and $a_x$ to $B$, then $B$ majorizes $A$. (Note $(x,y)\ne (i,i)$) and wlog $x\ne i$. By our claim, if $a_1<b_1<a_2<b_2<\cdots < b_{x-1}<b_x<a_x<a_{x+1}<\cdots<b_n$ then $A$ majorizes $B$, which is impossible by lemma 2 on $i,x$. Therefore we can use juggling (swapping adjacent elements) to justify that $A$ majorizes $B$ if and only if $a_i>b_i$. Note a juggle that changes the outcome must exist, from $a_1<b_1<a_2<\cdots<b_n$ to swapping $(a_j,b_j)$'s to get $b_1<a_1<\cdots<b_n<a_n$. The conclusion follows.

The answer is $100$. Let $n=100$. We claim that for a general positive integer $n$, the answer is $n$. Say $A\prec B$ whenever $B$ beats $A$. Construction: fix $1\le k\le n$. Define a rule $R_k$ that $A\prec B$ if and only if $a_k<b_k$. This is obviously a valid rule. We claim that we found all possible rules. Let $k\in\{1,2,\ldots,n\}$ be the minimal integer such that $$\{1,\ldots,k,k+n+1,\ldots,2n\}\prec\{k+1,\ldots,k+n\}.$$Observe that $k$ exists since $k=n$ gives $\{1,\ldots,n\}\prec\{n+1,\ldots,2n\}$. Moreover $k=0$ gives the false inequality $\{n+1,\ldots,2n\}\prec\{1,\ldots,n\}$ so $\{1,\ldots,k-1,k+n,\ldots,2n\}\succ\{k,\ldots,k+n-1\}$. We claim that $R_k$ holds. Let $A=\{a_1<a_2<\cdots<a_n\}$ and $B=\{b_1<b_2<\cdots<b_n\}$ be disjoint sets with $a_k<b_k$. Let $C=\{c_1<\ldots<c_n\}$ be a set of real numbers such that $C$ does not intersect $A,B$ and $$a_k<c_1<\ldots<c_n<\min(a_{k+1},b_k)$$where if $k=n$ then set $a_{k+1}=\infty$. Moreover, let $D=\{d_1<\ldots<d_n\}$ be a set of real numbers such that $D$ does not intersect $A,B$ and $$\max(c_n,b_{k-1})<d_1<\ldots<d_n<b_k$$where if $k=1$ then set $b_0=-\infty$. Since $$\{1,\ldots,k,k+n+1,\ldots,2n\}\prec\{k+1,\ldots,k+n\}$$we have $A\prec C$. Since $c_i<d_i$ we obtain $C\prec D$. Moreover, since $$\{1,\ldots,k-1,k+n,\ldots,2n\}\succ\{k,\ldots,k+n-1\}$$then $D\prec B$. This proves that $A\prec C\prec D\prec B$, so $A\prec B$.

Eyed wrote: Solved with Alan Bu, Chris Qiu, Edward Yu, Elliott Liu, Eric Shen, Isaac Zhu, Kevin Wu, Marvin Mao, Ryan Yang The answer is 0. Assume for the sake of contradiction that there exists a rule. Then, there must exist a deck of cards; which must be countable. However, the set of real numbers is uncountable, so such a deck can not exist. We conclude that no rules exist. Hi! In the statement it says infinite deck of cards. So they cannot be countable like the real numbers.

We claim the answer $100=n.$ We write $A\prec B$ if set $A$ beats $B.$ All sets in solution are supposed to be disjoint and sorted in increasing order. We claim, that all valid rules can be defined as follows: $\left \{ p_1,p_2,\dots ,p_n \right \} \prec \left \{ q_1,q_2,\dots ,q_n \right \}$ if and only if $p_k<q_k$ $(\circledast)$ holds for some fixed $k$. Clearly this works for $k=1,2,\dots,n.$ On the contrary, let's prove that for any rule, satisfying all three conditions, there exist $k$ satisfying $\circledast .$ Note that $$\mathcal M_i=\left \{ 1,2,\dots, i,i+1+n,i+2+n,\dots,2n\right \} \prec \mathcal N_i=\left \{ i+1,i+2,\dots ,i+n \right \}$$holds for $i=n;$ thus one may take $k$ as the minimal value of $i$ for which it holds. To prove that this satisfies $\circledast$, we need to show that for $\mathcal X=\left \{ x_1,\dots ,x_n\right \},$ $\mathcal Y =\left \{ y_1,\dots,y_n \right \}$ inequality $x_k<y_k$ implies $\mathcal X \prec \mathcal Y$ (then the converse implication also holds). Take arbitrary sets $\mathcal A=\left \{a_1,\dots ,a_n\right \},$ $\mathcal B=\left \{b_1,\dots ,b_n\right \},$ $\mathcal C=\left \{c_1,\dots ,c_n\right \}$, satisfying $$a_{k-1}<\min (x_1,y_1)<x_k<b_1<b_k<c_1<c_n<a_k<a_n<y_k<\max (x_n,y_n)<b_{k+1}.$$ For every $i$ we have $x_i<b_i,a_i<y_i$ which yields $\mathcal X\prec \mathcal B,\mathcal A\prec \mathcal Y$ by the second condtion. $\mathcal M_{k-1} \prec \mathcal N_{k-1}$ yields $C\prec A$ by the first condition, since elements of unions are sorted in the same way. $\mathcal N_k\prec \mathcal M_k$ yields $B\prec C$ by the first condition. All previous points imply $\mathcal X\prec \mathcal Y$ by the third condition, as desired $\Box$

Consider the general case of $n$ cards for each player. We claim there are $n$ such rules. Construction: The rule "A beats B iff $a_i > b_i$" for each $i = 1, ..., n$. This satisfies (1) since it only relies on the relative positions of the $i^{th}$ a's and b's. (2) is trivially true, and (3) by transitivity of $>$. Bound: We work in the relative positions of the reals instead of their values. Let $S(a, b)$ (abbreviated $S$) denote a string with $n$ a's and b's showing the relative increasing order of the cards. Let $g(S(a, b)) = \{i \mid a_i > b_i\}$ be a set of indices. Consider the following operation instead of (3): given $S_1(a, b)$ and $S_2(b, c)$, we construct $S_3(a, c) = S_1 * S_2$. Start with $S = S_1(a, b)$. For every $i$, for every c between $b_i$ and $b_{i+1}$ in $S_2(b, c)$, insert c in the position before the $b_{i+1}$ in $S$. For $i = n$, for every c after $b_n$, insert c at the end of $S$. Remove all b's from $S$ to obtain $S_3(a, c)$.

Note that given $S_1(a, b)$ and $S_2(b, c)$, by our construction of $S_3(a, c)$, we can find $A, B, C$ such that these relative orders are satisfied. Indeed, consider the final string $S$ without removing b's. This defines a relative order for $A, B, C$ and we can easily choose reals satisfying it. But note the pairwise relative orders are $S_1(a, b), S_2(b, c), S_3(a, c)$. Let $[a: k]$ denote a string of $k$ a's, and $[a]$ simply a string of a's. Define a 'majorising' string $S(a, b)$ where $g(S(a, b)) = \{1, ..., n\}$. Normalisation: Consider all $S(a, b)$ with $g(S(a, b)) = T$ for some fixed $T$. Then $S$ is 'normalised' if: \[S = \mid \underbrace{[a: k_1] [b: k_1]}_{i \not \in T} \mid \underbrace{t_1}_{i \in T} \mid \underbrace{[a: k_2] [b: k_2]}_{i \not \in T} \mid \underbrace {t_2}_{i \in T} \mid ...\]where $t_i(a, b)$ is majorising, and $i \in T$ means all indices are in $T$. Define a 'grand run' to be a run of consecutive indices either $\in T$ or $\not \in T$ such that indices immediately before and after this run lie in the opposite set. Claim 1: Let $\{e, e+1, ..., f\}$ be a grand run of consecutive indices in $T$, and $\{f+1, f+2, ..., g\}$ be the next grand run of consecutive indices not in $T$. Then any $S(a, b)$ with $g(S(a, b)) = T$ contains: \[... \mid \underbrace{[b_e, ..., a_f]}_{\substack{\textrm{starts $b_e$, ends $a_f$} \\ \textrm{elements are $b_e, ..., b_f, a_e, ..., a_f$} \\ i \in T \\ \textrm{a majorises b}}} \mid \underbrace{[a_{f+1}, ..., b_g]}_{\substack{\textrm{starts $a_{f+1}$, ends $b_g$} \\ \textrm{elements are $a_{f+1}, ..., a_g, b_{f+1}, ..., b_g$} \\ i \not \in T \\ \textrm{b majorises a}}} \mid ...\] Proof: Consider the block starting $b_e$, ending $a_f$ (necessary since $a_f > b_f > b_e$). Note that since $b_e < a_e < a_{e+1} < ... < a_f$, and $b_e < b_{e+1} < ... < b_f < a_f$, all desired a's and b's lie in the block. If the block also contains: $b_{e-1}$, then $b_e < b_{e-1}$; $a_{e-1}$, then since $e-1 \not \in T$ we have $b_e < a_{e-1} < b_{e-1}$; $a_{f+1}$, then $a_{f+1} < a_f$; $b_{f+1}$, then since $f+1 \not \in T$ we have $a_f > b_{f+1} > a_{f+1}$. Each is a contradiction. A similar argument establishes the $[a_{f+1}, ..., b_g]$ block. Majorising follows given this structure. $\square$ Claim 2: Let $S_3 = S_1 * S_2$. Then $g(S_3) \subseteq g(S_1) \cup g(S_2)$. Proof: Assume $\mu \not \in g(S_2)$ but $\mu \in g(S_3)$. Since $\mu \not \in g(S_2)$. \[S_2 = \underbrace{...}_{\textrm{contains $b_\mu$}} \underbrace{[c]}_{\textrm{contains $c_\mu$}} \underbrace{[b]}_{\textrm{starts $b_d$}} ...\]Thus, we have $d > \mu$. For $\mu \in g(S_3)$, we need: \[S_1 = \underbrace{...}_{\textrm{contains $b_d$}} \underbrace{[a]}_{\textrm{contains $a_\mu$}} ...\]otherwise if $a_\mu < b_d$ then $a_\mu < c_\mu$ contradiction. But now $b_\mu < b_d < a_\mu$ so $\mu \in g(S_1)$. $\square$ Claim 3: Let $R(a, b)$ be a string with $g(R(a, b)) = T$. Now define $R_0 = R, R_{i+1} = R_i * R_i$. Then we have $g(R_i) = T$ for all $i$, and for sufficiently large $N$, $R_N$ is normalised. Proof: Observe any grand run $\{\mu, ..., g\}$ not in $T$. \[R_i(a, b) = ... \mid \underbrace{[a:x_1]}_{\textrm{starts $a_\mu$}} \underbrace{[b:y_1]}_{\textrm{starts $b_\mu$}} \underbrace{[a:x_2]}_{\textrm{starts $a_{\mu+x_1}$}} \underbrace{[b:y_2]}_{\textrm{starts $b_{\mu+y_1}$}} ... \mid ...\]\[R_i(b, c) = ... \mid \underbrace{[b:x_1]}_{\textrm{starts $b_\mu$}} \underbrace{[c:y_1]}_{\textrm{starts $a_\mu$}} \underbrace{[b:x_2]}_{\textrm{starts $b_{\mu+x_1}$}} \underbrace{[c:y_2]}_{\textrm{starts $c_{\mu+y_1}$}} ... \mid ...\]By Claim 1, we have b majorising a, so $x_1 \ge y_1 \Rightarrow \mu + x_1 \ge \mu + y_1 \Rightarrow b_{\mu + x-1} \ge b_{\mu + y_1}$. Hence, the $[c:y_1]$ block gets inserted after the $[a:x_2]$ block under $*$, and \[R_{i+1}(a, c) = ... \mid \underbrace{[a:(x_1 + x_2)]}_{\textrm{starts $a_\mu$}} ... \mid ...\]Now note that if $\lambda \in g(R_i)$, $a_\lambda > b_\lambda > c_\lambda$ so $\lambda \in g(R_{i+1})$. Thus, $g(R_i) \subseteq g(R_{i+1})$. Yet by Claim 2, $g(R_{i+1}) \subseteq g(R_i)$ so $T = g(R_i) = g(R_{i+1})$. Eventually, in a $i \not \in T$-block, since $x_1 \to x_1 + x_2$ is a strict increase, due to a finite number of a's we eventually get that there is only one $[a]$-block. Yet for b to majorise a, we also have one $[b]$-block of the same size immediately after, hence normalised. $\square$

Now we work exclusively in normalised strings. Suppose we have $S_1, S_2$ both normalised, with $S_3 = S_1 * S_2$, and \[S_2(b, c) = ... \underbrace{[a]}_{\textrm{contains $c_\mu$}} \underbrace{[b]}_{\textrm{starts $b_d$}} ...\]. Claim 4: If $\mu \in g(S_2) \cap g(S_3)$, then $d \le \mu$, and $\exists d + j \in \{d, d+1, ..., \mu\}$ such that $d+j \in g(S_1) \cap g(S_2)$. Proof: By normalisation we have: \[S_2(b, c) = ... \mid \underbrace{... \underbrace{[c]}_{\textrm{contains $c_\mu$}} \overbrace{\underbrace{[b]}_{\textrm{starts $b_d$}} ...}^{\textrm{contains $b_\mu$}}}_{i \in g(S_2)} \mid ...\]hence $d \le \mu$. We necessarily also have: \[S_1(a, b) = \underbrace{...}_{\textrm{contains $b_d$}} \underbrace{[a]}_{\textrm{contains $a_\mu$}} ...\]If $d = \mu$, we also have $\mu \in g(S_1)$. Otherwise, since $\mu, d$ lie in the same $i \in g(S_2)$-block, we have $\{d, d+1, ..., \mu\} \subseteq g(S_2)$. Assume FTSOC that none of these lie in $g(S_1)$. Then they form a run of non-elements of $g(S_1)$, so actually by normalisation we have: \[S_1(a, b) = ... \mid \underbrace{\underbrace{[a]}_{\textrm{contains $a_d, ..., a_\mu$}} \underbrace{[b]}_{\textrm{contains $b_d, ..., b_\mu$}}}_{i \not \in g(S_1)} \mid ...\]a contradiction. $\square$ Claim 5: If $\mu \in g(S_1) \cap g(S_3)$, then either $\mu \in g(S_2)$, or $\mu < d$ and $d \in g(S_1) \cap g(S_2)$. In the latter case, we also have $d-1 \not \in g(S_2)$, $d-1 \in g(S_1)$. Proof: Assume $\mu \not \in g(S_2)$. Then: \[S_2(b, c) = ... \mid \underbrace{\underbrace{[b]}_{\textrm{contains $b_\mu$}} \underbrace{[c]}_{\textrm{contains $c_\mu$}}}_{i \not \in g(S_2)} \mid \underbrace{\underbrace{[b]}_{\textrm{starts $b_d$}} ...}_{i \in g(S_2)} \mid ...\]so $d > \mu$. We necessarily also have: \[S_1 = ... \mid \underbrace{\underbrace{...}_{\textrm{contains $b_\mu < b_d$}} \underbrace{[a]}_{\textrm{contains $a_\mu$}} ...}_{i \in g(S_1)} \mid ...\]Hence, $d \in g(S_1)$, and since $\mu \le d-1$, $d-1 \in g(S_1)$. In $S_2(b, c)$, the block containing $b_\mu$ must end $b_{d-1}$, so $d-1 \not \in g(S_2)$. $\square$ Claim 6: If $g(S_1) \cap g(S_2) = \O$, $g(S_1 * S_2) = \O$. Proof: By Claim 2, if $\mu \in g(S_3)$, then $\mu \in g(S_1)$ or $g(S_2)$. Note that in either case, because of Claims 4 and 5, there always exists an index in $g(S_1) \cap g(S_2)$, contradiction. $\square$ Let $W_a$ denote the set of winning strings $S(a, b)$ for $a$. Note that $S(a, b) \in W_a \iff S(b, a) \not \in W_a$, and let $S'(a, b) = S(b, a)$ be the complement of $S$. Note that $g(S') = \{1, 2, ..., n\} \setminus g(S)$. Furthermore, by (3), if $S_1, S_2 \in W_a$, then $S_1 * S_2 \in W_a$. Claim 7: Suppose $S \in W_a$, and $g(S) \subseteq g(P)$. Then $P \in W_a$. Proof: Assume FTSOC $P \not \in W_a$. Then $P' \in W_a \Rightarrow S * P' \in W_a$. Since $g(S) \subseteq g(P)$, $g(S) \cap g(P') = \O$. By Claim 6, $g(S * P') = \O \Rightarrow g((S * P')') = \{1, 2, ..., n\}$. Yet $(S * P')' \not \in W_a$, a contradiction of (2). $\square$ Take any $M \in W_a$ with smallest cardinality of $g(M)$. By (2), $|g(M)| > 0$. Claim 8: $|g(M)| = 1$. Proof: Assume FTSOC $|g(M)| > 1$. Take the maximal $\lambda \in g(M)$. Define $N$ to be any normalised string with $g(N') = M \setminus \{\lambda\}$. Hence, $\{\lambda\} = g(N) \cap g(M)$. Let $Q = N * M$. Note that by minimality, $N' \not \in W_a \Rightarrow N \in W_a$. Thus, $Q \in W_a$. Consider any $\mu \in g(Q)$. If $\mu \in g(M)$, then by Claim 4, $\exists d + k \le \mu$ such that $d + k \in g(M) \cap g(N)$. Thus, $d + k = \lambda$ necessarily. Now $\lambda \le \mu$, but by maximality $\lambda \ge \mu$, so $\lambda = \mu$. If $\mu \in g(N)$, then by Claim 5, either $\mu = \lambda$, or $\exists d > \mu$ with $d \in g(M) \cap g(N)$. In the latter, $d = \lambda > \mu$, and $\lambda-1 \not \in g(M), \lambda-1 \in g(N)$. Now we have $g(M) \cap g(Q) = \{\lambda\}$, and the max element in both being $\lambda$. At least one of $Q, M$ have $\lambda-1 \not \in G$, WLOG $Q$. Then we have $g(Q * M) \cap g(M) = \{\lambda\}$ and $g(Q * M) \cap g(Q) = \{\lambda\}$ by the same argument as above. Now $g(Q*M) = \{\lambda\}$, with $Q*M \in W_a$, contradiction. $\square$

Finally, take a $M \in W_a$ with $g(M) = \{\lambda\}$. Note that $W_a$ contains half of all strings. For every string $S(a, b)$, $\lambda$ lies in either $g(S)$ or $g(S')$. By Claim 7, every string $S(a, b)$ with $\lambda \in g(S)$ lies in $W_a$, which thus uniquely defines all elements of $W_a$. $\blacksquare$

There are exactly $100$ rules, all of which are described as follows: let $1 \leq k \leq 100$ be an integer and make $A$ beat $B$ iff $a_k>b_k$; these clearly work. I claim that these are the only rules. Call a pair of two sequences $A,B$ stairy if $\max\{a_i,b_i\}<\min\{a_{i+1},b_{i+1}\}$ for $1 \leq i \leq 99$, i.e. the intervals $[\min\{a_i,b_i\},\max\{a_i,b_i\}]$ are disjoint. Let the signature of a stairy pair $(A,B)$ be the set of $i$ such that $a_i>b_i$. If some signature results in $A$ beating $B$ then put it in a set $S$. Obviously the empty set is not in $S$. Claim: $S$ is closed under intersection. Proof: Let $s_1,s_2$ be two signatures in $S$. Starting from some sequence $A$, let $b_i=a_i+\varepsilon$ for $i \in s_1$ and $b_i=a_i-2\varepsilon$ otherwise. Then let $c_i=a_i+2\varepsilon$ for $i \in s_1 \cap s_2$, $c_i=a_i-\varepsilon$ for $i \in s_1 \oplus s_2$, and $c_i=a_i-3\varepsilon$ for $i \not \in s_1 \cup s_2$. If $\varepsilon>0$ is chosen to be sufficiently small then $(B,A),(C,B),(C,A)$ are all starry, and by definition $C$ will beat $B$ and $B$ will beat $A$. Thus $C$ will beat $A$, and $(C,A)$ has signature $s_1 \cap s_2$. $\blacksquare$ This is enough to imply that some signature $s$ is a subset of every element of $S$. I claim that $|s|=1$. Suppose otherwise, so we can write $s=t_1 \cup t_2$ where $t_1,t_2$ are disjoint nonempty sets. Starting from sequence $A$, let $b_i=a_i+\varepsilon$ for $i \in t_1$ and $b_i=a_i-\varepsilon$ otherwise. Let $c_i=a_i+2\varepsilon$ for $i \in t_2$ and $c_i=a_i-2\varepsilon$ otherwise. For sufficiently small $\varepsilon>0$ then $(A,B),(B,C),(A,C)$ are all starry, and by definition $A$ beats $B$, and $B$ beats $C$. But the signature of $(A,C)$ is $t_1 \cup t_2=s$, so $A$ should beat $C$: contradiction. Now consider some sequence $A$ and define a sequence as follows: let $A(0)=A$, then form $A(1)$ by increasing $A$'s $k$-th element and decreasing the rest of the indices by a tiny amount. Then form $A(2)$ by increasing $A(1)$'s $k$-th element and decreasing the rest of the elements by a tiny amount again, except that we make $a(2)_{k-1}$ less than $a(0)_{k-2}$. Then form $A(3)$ by increasing $A(2)$'s $k$-th element and decreasing the rest of the elements by a tiny amount again, except we make $a(3)_{k-1}$ and $a(3)_{k-2}$ less than $a(0)_{k-3}$. Repeat until we reach some $m$ where $a(m)_{k-1}<a(0)_1$, and then do an analogous thing to elements $k+1,k+2,\ldots$. Since $A(i)$ beats $A(i-1)$ for all $i$, we end up finding that $A$ loses to the sequence $A'$ where $a'_k>a_k$ but $a'_{k-1}<a_1$ and $a'_{100}<a_{k+1}$. Now for any sequence $B$ such that $b_i>a'_i$ for all $i$ we find that $B$ beats $A'$ and therefore beats $A$. Moreover, across all $B$, it's clear that any relative ordering of $a_1<\cdots<a_{100}$ and $b_1<\cdots<b_{100}$ can be achieved so long as $a_k<b_k$, hence we find that the corresponding rule is precisely that where $A$ beats $B$ when $a_k>b_k$. $\blacksquare$ Remark: I believe drawing out the sequences and doing everything visually before transforming to this more algebraic language is very helpful. This solution is a bit less direct than the other one but it feels more satisfying: you maybe get more of a sense of what's actually going on.

I recently wrote something about this problem. It might be interesting.

The answer is $100$. Write $X\succ Y$ if $X$ beats $Y$. Let $A = \{1, \dots, 100\}$. Let $B_k = \{1.5, 2.5, \dots, k - 0.5\} \cup \{k - 0.25\} \cup \{k + 1.5, k + 2.5, \dots, 100.5\}$. Case 1: $B_k \succ A$ for all $k$ Claim: $\{1-1000, 2-1000, \dots, i-1000\} \cup \{i + 1.5, i + 2.5, \dots, 100.5\} \succ A$ Proof: We will prove this by induction on $i$. In the base case, by definition $\{1.5, 2.5, \dots, 100.5\} \succ A$. Assume $\{1-1000, 2-1000, \dots, i-1000\} \cup \{i + 1.5, i + 2.5, \dots, 100.5\} \succ A$. Then, since $B_{i+1}\succ A$, we have \[\{1.5-1000, 2.5-1000, \dots, i + 0.5-1000, i+1-1000\} \cup \{i + 2.5, \dots, 100.5\}\succ \{1-1000, 2-1000, \dots, i-1000\} \cup \{i + 1.5, i + 2.5, \dots, 100.5\}.\]Therefore, \begin{align*} \{1.5-1000, 2.5-1000, \dots, i + 0.5-1000, i+1-1000\} \cup \{i + 2.5, \dots, 100.5\} &\succ A\\ \{1-1000, 2-1000, \dots, i+1-1000\} \cup \{i + 2.5, \dots, 100.5\} &\succ A. \end{align*} Therefore, $\{1-1000, 2-1000, \dots, 100-1000\}\succ a$, a contradiction. Case 2: $B_k \prec A$ and $B_j \prec A$ for $k< j$ Then, we have \begin{align*} \{0.5, 1.5, \dots, k - 1- 0.5\} \cup \{k+0.01\} \cup \{k+1- 0.5, \dots, 100-0.5\} &\succ A\\ \{0+0.01, 1+0.01, \dots, k-2+0.01\}\cup \{k-1+0.01\} \cup \{k+0.01,\dots, j-2+0.01\} \cup \{j-0.5+0.01\}\cup \{j+0.01, \dots, 99+0.01\} &\succ \{0.5, 1.5, \dots, k - 1- 0.5\} \cup \{k+0.01\} \cup \{k+1- 0.5, \dots, 100-0.5\}\\ \{0+0.01, 1+0.01, \dots, j-2+0.01\} \cup \{j-0.5+0.01\}\cup \{j+0.01, \dots, 99+0.01\} &\succ A, \end{align*}a contradiction. Therefore, there is exactly one $k$ such that $B_k\prec A$. If we start with a set of 100 cards and slightly increase every card except for the $k$th card, which we slightly decrease, we get a new set that loses to the original set. By repeating this process, for arbitrarily small $\epsilon$ and arbitrarily large $N$, we get \[\{k-(k-1)\epsilon, k-(k-2)\epsilon \dots, k-\epsilon\}\cup \{k-\epsilon\} \cup \{k+1+N, \dots, 100+N\} \prec A.\] Assume we have a set $c_1<c_2<\dots <c_{100}\in C$ disjoint from $A$. Assume $c_k<k$. Then, for sufficiently small $\epsilon$ and sufficiently high $D$, we have \begin{align*} \{k-(k-1)\epsilon, k-(k-2)\epsilon \dots, k-\epsilon\}\cup \{k-\epsilon\} \cup \{k+1+N, \dots, 100+N\} &\succ C\\ C &\prec A. \end{align*}So, the winner is whichever set has a higher $k$th smallest value. Clearly all rules of this form work. So, there are 100 rules.

The answer is $\boxed{100}$. In general, if we replace $100$ with $n$ and ($200$ with $2n$), the answer is $n$. There is one rule for each $k$ where $A$ beats $B$ iff $a_k>b_k$; clearly these work. Unless noted otherwise, assume that every $\varepsilon$ used in the solution is sufficiently small. When I call a set "greater than" another set, I mean that the former set "wins against" the latter set. Fix a set $A$. For positive integers $k\le n$, define $B_k$ to be the set $\{b_1,b_2,\dots,b_n\}$ with $b_i=a_i-\varepsilon$ if $i\ne k$ and $b_k=a_k+\varepsilon$. The main claim is that $A$ is less than exactly one set out of $B_1,B_2,\dots,B_n$. First, note that if it is greater than all of these, then \[\{a_1,a_2,\dots,a_n\}>\{a_1+n\varepsilon,a_2-\varepsilon,\dots,a_n-\varepsilon\}>\{a_1+(n-1)\varepsilon,a_2+(n-1)\varepsilon,\dots,a_n-2\varepsilon\}>\dots>\{a_1+\varepsilon,a_2+\varepsilon,\dots,a_n+\varepsilon\},\]contradiction. But if $A$ is less than two of the $B_i$, WLOG say they are $B_1$ and $B_2$, then \[\{a_1,a_2,\dots\}<\{a_1+\varepsilon,a_2-2\varepsilon,\dots\}<\{a_1-\varepsilon,a_2-\varepsilon,\dots\},\]contradiction. Now let $k$ be the value such that $B_k$ is greater than $A$. Then we claim that the rule is simply comparing $a_k$ and $b_k$. To do this, we will prove that if $a_k<b_k$, then $A<B$. Set $\delta$ to be sufficiently small even compared to $\varepsilon$. And let $N$ be a really large number. Indeed, \begin{align*} & \{a_1,a_2,\dots,a_{k-1},a_k,a_{k+1},a_{k+2},\dots,a_n\} \\ <& \{-N+\varepsilon,a_1+\varepsilon,\dots,a_{k-2}+\varepsilon,a_k+\delta,a_{k}+\varepsilon,a_{k+1}+\varepsilon,\dots,a_{n-1}+\varepsilon\} \\ <& \{-N+\varepsilon,-N+2\varepsilon,\dots,a_{k-3}+2\varepsilon,a_k+2\delta,a_{k}+\varepsilon,a_{k}+2\varepsilon,\dots,a_{n-2}+2\varepsilon\} \\ <& \dots \\ <& \{-N+\varepsilon,-N+2\varepsilon,\dots,-N+(k-1)\varepsilon,a_k+(k-1)\delta,a_k+\varepsilon,a_k+2\varepsilon,\dots,a_k+(n-k)\varepsilon\} \\ <& \{b_1,b_2,\dots,b_{k-1},b_k,b_{k+1},b_{k+2},\dots,b_n\}\;\blacksquare \end{align*}