Here's a "bash" to do this. The answer is $2n-2.$ A construction can be given by all $n$ subsets of size $n-1$ and any $n-2$ singletons. Before proceeding to the real problem, we make a few definitions. Let $T$ be a collection of subsets of $X$ and $|T| \ge 2n-1.$ We show that there must be a subcollection of $T$ that is tight. A note: we also $T$ to contain a set multiple times, but not the empty or full set. Say that an element $x \in X$ is compressible to an element $y \in X$ if $\forall S \in T, x \in S \implies y \in S.$ In other words, for all sets $S$ in $T$, if $x$ is in $S$, then $y$ is also in $S$. By abuse of notation, say that $T$ is compressible if there exist some $x \neq y \in X$ such that $x$ is compressible into $y.$ Now we make a slightly contrived definition (but will be needed for the proof). Say that $T$ is irreducible if for all $T' \subseteq T$ with $|T'| = |T|-1$, $T'$ is not compressible. In other words, $T$ is irreducible if its impossible to delete a single set of $T$ to make it compressible. We now prove the following lemma. Lemma: If $T$ is irreducible, then we can choose $A_1, A_2, \dots, A_k \in T$ such that they are tight, and $|A_1 \cup \dots \cup A_k| = n-1.$ Proof: We proceed by induction on $n$. Let $B = \{S \in T : n \in S. \}.$ Define $B' = \{C \backslash \{n\} : C \in B \}.$ So $B'$ is the sets in $B$, but removing the element $n$. Consider the multiset $T' = ((T \backslash B) \cup B') \backslash \{ \emptyset, [n-1] \}.$ Note that $|T'| \ge 2n-3,$ and $T'$ only consists of subsets of $[n-1].$ Also, if $T$ is irreducible, then so is $T'.$ Indeed, the sets $[n-1], \emptyset$ don't affect irreducibility, and neither does deleting the element $n$. By induction, we can find some $S_1, S_2, \dots, S_k \in T'$ that are tight and $|S_1 \cup \dots \cup S_k| = n-2.$ WLOG, say that $S_1 \cup \dots \cup S_k = [n-2].$ To complete the induction, we simply have to add in some more sets containing $n$ to get a total union of size $n-1.$ Out of the $S_1, \dots, S_k$, some came from $T \backslash B$ and some came from $B.$ If $\ge 2$ came from $B$, then we're done, $n$ was in the union of the $S_i$ extended back to $[n].$ If $0$ or $1$ came from $B$, we need to add more in. If in $T$, there are $\ge 2$ sets containing $n$ but not $n-1$, then we're done, since we can just use those too. Otherwise, $\le 1$ sets in $T$ contain $n$ but not $n-1$. This contradicts irreducibility or $T$ though, as we can just delete the sets containing $n$ but not $n-1$, and then compress. This completes the proof. Now, to finish the problem, we simply have to deal with sets that are reducible (not irreducible). We do this part by induction on $n$. Take $T$, and say that we can delete a set $S \in T$ and then compress $x \to y.$ What I mean by ``compress" if to delete the element $y$ from all sets in $T$. Let $T'$ be this new collection of subsets of $(X \backslash \{y\})$.This could result in a single empty set, so when we do this, the number of sets decreases by at most $2$ (one for $S$, one more for empty set). Now, by induction we can find a tight subcollection in $T'.$ Now, as before we can extend this back to $X$. Say that we have used $S_1, \dots, S_k \in T.$ If $x$ is used in at least $2$ of the $S_i$, we're done, since $y$ will also have been used in $2$ of the $S_i$ (because we were able to compress $x \to y.$) By induction, $x$ won't be used exactly once. If $x$ is used zero times, then we might run into an issue because its possible that $y$ is in exactly one of the $S_i$, but $x$ is in none of them. All other cases are fine (if $y$ is used $0$ or $\ge 2$ times). If there are at least $2$ sets in $T$ that have $Y$ but not $X$, then we can just use both and finish. Otherwise, at most one set has $Y$ but not $X$. Also, by the reducibility of $T$, at most one set in $T$ has $X$ but not $Y$. Now delete both these sets from $T$. In all remaining sets in $T$, either both $X, Y$ are there or neither. So they are equivalent elements! So they can be treated at the same element. Therefore, we can reduce down to $|X| = n-1$, and we are done by induction.

The answer is $2n-2$. Here is the solution communicated to me by the American team. For convenience, let $X = \left\{ 1, \dots, n \right\}$. An example of a construction is to take all sets of size $n-1$ and any $n-2$ singleton sets. We proceed by induction on $n \ge 2$ with the base case being immediate. Consider such a collection $\mathcal C$ of sets. We will assume that for any two elements $i$ and $j$, there is a set containing one but not the other, since otherwise we could merge the two together and then induct down. Assume WLOG that $1$ appears the fewest number of times. We'll partition $\mathcal C$ into two parts: the sub-family $\mathcal F$ of sets which contain $1$, and the sub-family $\mathcal G$ which are subsets of $\{2, \dots, n\}$. Then we contend that: Claim: Each of $2$, $3$, \dots, $n$ appears at least once in $\mathcal G$. Proof. Assume for contradiction that any time $2$ is present, so is $1$. By the merging argument, there is also a set containing $1$ but not $2$. So there are strictly more $1$'s than $2$'s, which is absurd. $\blacksquare$ Now, since $\bigcup \mathcal G = \{2, 3, \dots, n\}$ it follows that some element, say $2$, is contained in exactly one element of $\mathcal G$, say $g$. (Otherwise $\mathcal G$ is a tight collection.) Now we remark that there is at most one set $f$ in $\mathcal F$ containing $1$ but not $2$: (since we assumed that $1$ appears at most as many times as $2$, and yet $2$ appeared only once in $\mathcal G$, so it must appear in all but one set of $\mathcal F$). Hence we delete $f$ and $g$ from $\mathcal C$, and in this case the merging argument now applies, allowing us to induct down.

v_Enhance wrote: Claim: Each of $2$, $3$, \dots, $n$ appears at least once in $\mathcal G$. You don't really need this; clearly $\mathcal G$ is nonempty, and the union $\bigcup \mathcal{G}$ does not contain 1, so you can immediately say $\mathcal G$ would be tight if there is not some element appearing in $\mathcal G$ exactly once. From here, the reduction works the same way.

I guess this was too easy for a RMM 3.

Not really i found this to be beautiful.

Remark: This felt immensely satisfying. For simplicity let the $n$ element set $X = \{1, 2, \ldots , n\}$. I claim that the answer is $\boxed{2n-2}$. This can easily be constructed by taking all $n$ subsets of $X$ with cardinality $n-1$ along with some $n-2$ single element subsets; if the subcollection contains two sets of cardinality $n-1$ then it is already not tight since the union is $X$, and if not, there will always be an element that lies in only the $n-1$ element subset chosen since the described collection only has $n-2$ one element sets. We will complete the rest of the problem by trying to induct down. The base case of $n = 2$ is clear. Next, we consider a collection $\mathcal{C}$ of sets that satisfies the problem conditions. We will make an optimal assumption: for any two elements $a, b \in X$, there is some set $\in \mathcal{C}$ containing only one of them. If not, then we can combine $a$ and $b$ into one element and reduce $|X|$, since we only care about the number of sets in $\mathcal{C}$ and this reduction does not affect that. Consider some number, say $1$, that appears in the least number of sets in $\mathcal{C}$ (clearly exists). We split the colection $\mathcal{C}$ into two subcollections, $\mathcal{A}$ and $\mathcal{B}$, where the former consists of sets containing $1$ and the latter consists of sets not containing $1$. There must be some element in $X \symbol{92} \{1\}$ that appears in only one set in $\mathcal{B}$, or else $\mathcal{B}$ would be tight, since its union is already guaranteed to be a proper subset of $X$. Suppose that this element is $2$ and the unique set it appears in is $b \in \mathcal{B}$. Now we consider subcollection $\mathcal{A}$, where each set in $\mathcal{A}$ contains $1$. Note that there can be at most one set in $\mathcal{A}$ not containing $2$, or else the number of sets containing $1$ exceeds the number of sets containing $2$, a contradiction to our previous assumption of minimality. Let this set, if it exists, be $a \in \mathcal{A}$. Next we set aside $a$ and $b$. Among the remaining sets in $\mathcal{C}$, all sets either contain both $1$ and $2$ or none of $1$ or $2$. In this case, we can combine $1$ and $2$ into a single element, therefore reducing $|X|$ by one, deleting $\leq 2$ sets in $\mathcal{C}$. It is possible to keep repeating this operation until $|X|$ reduces to $2$, and we see that therefore $|\mathcal{C}| \leq 2(|X|-2) + 2 = 2|X| - 2 = 2n-2$ as desired. $\blacksquare$

jj_ca888 wrote: Remark: This felt immensely satisfying. Same

Solved with tapir1729. We claim that the answer is $2n-2$. As a construction, consider all subsets of size $n-1$, and all but $2$ of the singletons. Now, we proceed with induction. The claim is trivial for $n=2$, so we jump to the inductive step. Consider the family, $\mathcal{F}_k$ of subsets which contains all the $A_i$ which don't contain $k$. So, $k\not\in \bigcup\mathcal{F}_k$, so the union is not $X$. Hence, there must be an element of $\mathcal{F}_k$ which contains an element, $j$, not contained by any other element of $\mathcal{F}_k$. In particular, this means that there exists a pair $(k,j)$ for any $k$ such that there's a unique set which contains $j$ but not $k$. Using this, we can construct a directed graph on the elements of $X$, where we have $k\to j$ if and only if there's exactly one set which contains $j$ but not $k$. As each vertex has outdegree at least one, we can choose a cycle, $\mathcal{C}$ from our graph. Let the family of sets associated with $\mathcal{C}$ be denoted $S(\mathcal{C})$, and the elements associated with it be $v(\mathcal{C})$. Now, for any set not in $S(\mathcal{C})$, it is easy to see that all elements of $v(\mathcal{C})$ are in the same state (ie either all in the set or all out of it), or else uniqueness is violated. Hence, we can condense them all elements of $v(\mathcal{C})$ into a single element $v'$. Namely, if $v(\mathcal{C})\in S$, we replace it with $S\setminus v(\mathcal{C})\cup \{v'\}$. Using our inductive hypothesis on $X\setminus v(\mathcal C)\cup \{v'\}$, we see that $\{A_i\}\setminus S(\mathcal C)$ has at most $2((n-|\mathcal C|+1)-1)=2(n-|\mathcal C|)$ elements, so $\{A_i\}$ has at most $$2(n-|\mathcal C|)+|\mathcal C|=2n-|\mathcal C|$$elements. A cycle has at least 2 vertices, so this gives our bound of $2n-2$, as desired.

Here is a solution which interprets "tightness" in a much more natural way IMO. Very nice problem! Refer to a subset $S$ of $X$ by an $n$-digit binary string in which the $i$th digit is $1$ if $S$ contains the $i$th element of $X$ and $0$ otherwise. The answer is $\boxed{2n-2},$ achieved by the set of all strings with exactly $n-1$ ones and all but two strings with exactly $1$ one. To show this, we will induct on $n.$ For the base case $n=2$, just note that there are only $2$ proper non-empty subsets of a two-element set. Now assume the claim is true for $n=k-1,$ and suppose FTSOC there exists a collection of $2k-1$ subsets of a $k$-element set such that no subcollection is tight. Arrange the binary strings associated with the subsets in a $k\times (2k-1)$ array. The key claim is the following. $\textbf{Claim: }$ There exist two columns which coincide in at least $k-2$ rows. $\emph{Proof: }$ WLOG suppose that column $1$ contains the maximum number of $0$s, say $m,$ and the first $m$ rows start with $0.$ Since the set of the first $m$ strings isn't tight, there is some column, WLOG column $2,$ which contains exactly $m-1$ zeroes in the first $m$ rows. But column $2$ contains at most as many $0$s as column $1,$ so column $2$ contains at least $k-m-1$ ones in rows $k+1,k+2,\dots,m.$ Therefore, columns $1$ and $2$ coincide in at least $k-2$ rows, as desired. $\blacksquare$ Now, viewing the two columns as one gives an immediate contradiction by the inductive hypothesis.

One of the most interesting problem I've ever seen the answer is $2n-2$ suppose $X=\{1,2\dots n\}$ for $2n-2$ consider all sets with cardinality $n-1$ and the sets $\{1\} ,\{2\} \dots \{n-2\}$ now suppose $|X|\ge 2n-1$ call a set such $X$ Arabic let $X$ be the Arabic set with maximal cardinality and induct on $n$ now the main step is: consider a directed graph $\mathbb{G}$ with vertices represents $\{1,2 \dots n\}$ and an edge from $i$ to $j$ if and only if : every set contains $i$ also contains $j$ except exactly one set $S_{(i,j)}$ if $outdeg(v_i)=0$ call it stingy now we claim(1): there's only one stingy set (Actually all Arabs are generous ) proof: suppose there exists two $a,b$ now one of the sets $\{a\} , \{b\}, \{a,b\}$ is not contained in $X$ WLOG let it $\{a,b\}$ now we have $\{X+\{a,b\}\} $ is not Arabic so there exists a subset $S$ of $X$ such that it has only $a$ and two or more some elements and no $d$ now if there exists a subset $H$ has $d$ without $a$ contradiction since $S+H$ will be tight so $a$ is not stingy contradiction $\blacksquare$ claim(2): $G$ has a cycle proof: suppose not now obviously there exist $v$ with $indeg(v)\ge 2$ let it $3$ and two beyond it $1,2$ let $T$ be the set of $v_i$ such that there exists a path from $3$ to $v_i$ we can see that $1 $ not in $T$ as claim(1) there suppose WLOG $\{T+\{1\}\}$ is not in $X$ then there exists $z $ not in $\{T+\{1\}\}$ and $i \in \{T+\{1\}\}$ such that $\vec{v_iv_z}$ is in $\mathbb{G}$ so there exists a cycle $\blacksquare$ now if there exists a cycle $\mathbb{C} \in \mathbb{G}$ let its vertices be $v_1,v_2,\dots v_c$ let $X'=X -S_{(v_i,v_{i+1})}$ for $i \in \{1,2,\dots c\}$ and let $Y=\{1,2\dots c\}$ so the elements of subsets of $X'$ are $\{Y,c+1,c+2 \dots n\}$ so by induction hypothesis we're done and eventually we win

Wow, this is so good. The answer is $2n-2$, for simplicity say the elements of $X$ are $1,2,\cdots,n$ and the set in question is $S$. The construction then is to take all singletons except $1$ and all $n-1$ element sets except $\{2,3,\cdots,n\}$. Consider all elements of $S$ that don't contain $n$. If this is empty, then we can directly induct down so assume $S$ is nonempty. Because the union of these elements is not tight, and since it cannot equal $X$, there is an element, say $i$ that occurs in exactly one such element. Suppose we delete this element as well as $\{1,2,\cdots,n\} \setminus \{i\}$, if it exists in $S$. We're left with $|S| - 2$ subsets, and replacing every occurrence of $i$ with $n$ as well (and delete repetitions inside a set). Then by the inductive hypothesis, we get that $|S| - 2 \leqslant 2n-4 \implies |S| \leqslant 2n-2$, as desired.

Answer is $2n - 2$. We will induct on $n$ with our base cases being true. First note that there is some set that contains only $p$, not just $q$. We can split the subsets $A_i$ into two groups, one that only contains all of the subsets that have $p$. The union of all of the subsets in the second group is a proper subset of $X$ and thus must contain exactly one subset that has $q$(that does not have $p$), so proven. Then WLOG let $a$ be the element that is used the least amount of times in $A_i$. Split the subsets into two groups, $G_1$ and $G_2$ with $G_1$ being the set of subsets containing $a$ and vice versa. Then since the union of all sets in $G_2$ is not tight then there must be some element $a$ that only shows up once in the subsets in $G_2$. It follows from the minimality of $e$ that there is exactly one subset in $G_1$ that only contains $a$(not $e$) but the rest of the subsets must contain $e$. Then we can delete the two subsets that contain exclusively $a$ and $e$ to induct down. Then the rest of the subsets that contain either $a$ or $e$ contain both of them, so we merge the two elements together to create $æ$(inducting from $n$ elements to $n-1$ elements), so we are done. forgot to add that this is a really good problem, maybe one of my favorites!

We claim our answer is $\boxed{2n-2}$. We inductively construct this collection by using our base case $n=2$ with $\{\{1\},\{2\}\}$. For each $n=k \ge 3$, take the collection from $n=k-1$, add $k$ to each set, and finally add the sets $\{1,2,\ldots,k-1\}, \{k\}$. We again use induction to prove maximality. Let the collection be $S$, and suppose WLOG the frequency of $k$ is the lowest. Separate $S$ into two sets $S_1$, which contain $k$, and $S_2 = S \setminus S_1$. Note that we must have some element $\ell \in \{1,2,\ldots,k-1\}$ which appears exactly once in $S_2$, or $S_2$ is a tight collection. Using the minimality of $k$, it follows there is also at most one set in $S_1$ which contains $\ell$ but not $k$. If there were no sets for which this is the case, we can essentially merge $\ell$ and $k$, allowing us to induct downwards. Otherwise, we remove these two sets from $S$, and we have a lower case again by merging. $\blacksquare$

The answer is $\boxed{2n-2}$. WLOG suppose that $X = \{1,2,\dots,n\}$. Consider the following sets: \[A_i = \{i\}, \quad 1 \le i \le n,\]\[B_i = \{1,2,\dots,i\}, \quad 1 \le i \le n-1.\] A valid construction consists of the union of all the $A_i$ and $B_i$, which one can easily confirm. To show the bound, we proceed by strong induction on $n$. The base case $n=2$ can be checked by hand. For the inductive step, given that $|X| = k$, we will reduce down to $k-1$. In a maximal tight subcollection $S$, denote the element that appears the least be $a$. Split $S$ into two collection of sets $S_1$, which contains all the sets that contain $a$, and $S_2 = S \setminus S_1$. In $S_2$, there must be an element $b \in X \setminus \{a\}$ that appears exactly once to prevent $S_2$ from being tight; denote that set as $A$. However, by our minimality condition on $a$, there is at most one element in $S_1$ such that $b$ doesn't appear in it, denoted as set $B$. Hence, we may delete the $A$ (if present) and $B$ to get \[|S| \le |S \setminus \{A,B\}| +2 \le (2k-4)+2 = 2k-2,\] as desired.