In this proof, a residue class will refer to a residue class modulo 47. For integers $i$ with $1 \leq i \leq 46$, let $S_i = \{x \in X : x \equiv i \pmod{47} \}$. For any nonzero residue class $i$, we define the weight of $i$, denoted by $w(i)$, to be the cardinality of $S_i$. (Then trivially, $w(1) + w(2) + \cdots + w(46) = 10,000$.) In addition, we call a set of integers $A$ good if $3A \cap 2A = \emptyset$, where $3A = A+A+A$ and $2A = A+A$ (where we are taking sumsets modulo 47.) Our problem becomes equivalent to finding some good set $Y$ such that the sum of the weights of all residues in $Y$ is at least 2010. We claim that we can find a good $A = \{a_1, a_2, \ldots, a_{10}\}$ with 10 nonzero residue classes. Indeed, we may let $a_i = 18 + i$. $2A = \{38, 39, \ldots, 56\} =\{0, 1, 2, \ldots, 9, 38, 39, \cdots, 46\}$ and $3A = \{57, 58, \ldots, 84\} = \{10, 11, 12, \ldots, 37\}$, where equality is taken modulo 47. $A$ is clearly good, so our claim is proven. It is trivial to see that if $A$ is good, then $kA = \{ka \pmod{47} : a \in A\}$ must be good as well. Let $\sigma(kA)$ be the sum of the weights of the elements of $kA$. I claim that we can find some $k$ such that $\sigma(kA) \geq 2007$, which clearly will complete our proof. Suppose for the sake of contradiction that for $k$ between 1 and 46, $\sigma(kA) \leq 2006$. Noting that $a_j \{1, 2, \ldots, 46\} = \{1, 2, \ldots, 46\}$ modulo 47, we have \begin{align*} 2006 \cdot 46 &\geq \sigma(A) + \sigma(2A) + \cdots + \sigma(46A) \\ &= \sum_{i=1}^{46} \sigma(iA) \\ &= \sum_{i=1}^{46} \sum_{j=1}^{10} w(ia_j) \\ &= \sum_{j=1}^{10} \sum_{i=1}^{46} w(ia_j) \\ &= \sum_{j=1}^{10} \sum_{i=1}^{46} w(i) \\ &= \sum_{j=1}^{10} 10,000 \\ &= 100,000 > 92276 = 2006 \cdot 46, \end{align*} which is a contradiction.

Zhero wrote:

Furthermore, we basically have to find such a set $A$ to optimally bound the case in which the residues in $X$ are approximately evenly distributed. These ideas have also appeared before in USA TST 2001.3 (a special case of Erdős's more general result).

Alternatively, $A=\{ 1, 3, 5, 7, 9, -1, -3, -5, -7, -9 \}$ $m=a-b+c-d+e$ is nonzero because it is odd. Then, if 47 divides $m$ $45\geq\sum \left | a \right |\geq\left | m \right |\geq47$ contradiction

I will speak in $\mathbb{Z}_p$ where $X,Y$ are multisets and $p=47$ is prime. Let $A$ be the middle-fifth of $\{1,...,p\}$, namely $\{19,...,28\}$. Notice $|A|=10$. Also, $A+A$ is the first and last fifths and $A+A+A$ is the second, third and fourth fifths, so $A+A \cap A+A+A = \varnothing$. For any constant $c$ if $Y=X \cap Ac$ then $Y$ will work, and therefore $|X \cap Ac| \le 2006$ for any $c$. For any $x \in X$ let $f(x) = \{ c \in \mathbb{Z}_p : cx \in A\}$, and notice $|f(X)|=|A|$. Then if we pick $c \in \mathbb{Z}_p$ at random, $2006 \ge E[ |X \cap Ac| ] = \sum_{x \in X} \text{Pr}[c \in f(x)] = |X| \left( \frac{|A|}{p-1} \right) = 10000*10/46 \approx 2173 > 2006$, Q.E.D.

Clarification of question: Is it for $a, b, c, d, e \in Y$ pairwise distinct or possibly equal? Thanks!

For $i \neq 0 \mod 47$, let $a_i$ denote the number of elements congruent to $i$ modulo $47$ in $X$. Let $t$ be a primitive root modulo $47$. Let $S_k = \{19t^k, 20t^k, ..., 28t^k\}$, $0 \le k \le 45$. Note $S_k$ satisfies the condition $a-b+c-d+e \neq 0 \mod 47$ for $a, b, c, d, e \in S_k$. If $T_k = \sum_{i \in S_k}a_i \ge 2007$ for some $k$, then we're done. Suppose the contrary holds. Then $92276 = 2006*46 \ge T_0 + T_1 + ... + T_{45} = 10(a_1+a_2+...+a_{46}) = 10*10000 = 100000$, a contradiction, and we're done.

It seems my construction was a bit convoluted . Consider the following $10$-element subset of elements of $\mathbb{F}_{47}$: \[A=\{1,6,\dots , 41,46\}.\]We claim that for any $a,b,c,d,e$ in this subset, we have $47\nmid a-b+c-d+e$. It is equivalent to show that $(A+A)\cap (A+A+A)=\emptyset$. Note that $A+A$ consists of the numbers in the arithmetic sequence $2,7,\dots , 92$ reduced modulo $47$ and $A+A+A$ consists of the numbers in the arithmetic sequence $3,8,\dots , 138$ reduced modulo $47$. Note that the differences of terms are always some element of the arithmetic sequence $136,131,\dots , 46$ reduced modulo $47$ or the negative version thereof. However, no multiples of $47$ are contained in the first sequence, since $47\equiv 2\pmod{4},94\equiv 4\pmod{4}$, but the first sequence consists of the numbers congruent to $1\pmod{5}$ between $46=0+46$ and $136 = 141-5$. Similarly, no multiples of $47$ are contained in the other sequence, hence we are done. Now, for any $j\in\mathbb{F}_{47}$, define \[A_j = \{aj\mid a\in A\};\]as long as $j\ne 0$, this new set also has that for any $a,b,c,d,e$ contained within it, $47\nmid a-b+c-d+e$. Now, pick a random set $A_j$. Any of the 10 elements of $A_j$ is congruent to $10,000/46$ elements of $X$ on average, hence any random set $A_j$ constitutes the congruency classes of $100,000/47 \approx 2173$ elements of $X$. Hence, there must exist some $A_j$ constituting the congruency classes of at least $2007$ elements of $X$, so then we can just pick the $x\in X$ in those congruency classes and find the set we need.

Define $S_k=\{19k,20k,\ldots 28k\}$ mod 47. Let $Y_k$ be the number of elements of $Y$ in $S_k$. Note that $Y_k$ satisfies the $a-b+c-d+e$ property. Then, by double counting \[\sum S_k = 10\cdot 1000\]so for some $S_k$ its size is $\geq \frac{10\cdot 10000}{46} > 2007$ and we are done.

There exists a number mod $47$ that appears at most $250$ times. By multiplying every number by a number, we can assume without loss of generality at most $250$ numbers in $X$ are $16$ mod $47$. Consider the set $S_1=\{1,6,11,16,21,26,31,36,41,46\}$. Then, for any five elements $a,b,c,d,e$ in this set, $a-b+c-d+e$ is $1$ mod $5$. It is also between $-89$ and $136$, so it cannot be a multiple of $47$. Now, let $S_i=\{i,6i,11i,16i,21i,26i,31i,36i,41i,46i\}$. Then, each nonzero residue appears in exactly $10$ of $S_1$, $S_2$, $\ldots$, $S_{46}$. Since there are $10000$ integers in $X$, there exists a set such that at least $10000\cdot\frac{10}{47}$ elements in $X$ are equivalent to some number in $S_i$ mod $47$. Since $10000\cdot\frac{10}{47}>2127>2007$, this means that there exists a $2007$-element subset of $X$ that satisfies the conditions.

Suppose otherwise. Let $c_i$ be the number of elements $x$ in $X$ such that $x\equiv i\pmod{47}$, where $1\le i\le 46$. Call a set good if it has only elements from $1$ to $46$, inclusive, and satisfies the condition in the problem. Suppose we have a good set with elements $a_1, a_2, \dots, a_n$. Then note that $c_{a_1}+c_{a_2}+\dots+c_{a_n}<2007$. We can multiply all the elements in this set by any number from $1$ to $46$, inclusive, and reduce $\bmod \hspace{0.1cm} 47$ to obtain another good set. This logic gives that $n(c_1+c_2+\dots+c_n)=10000n<2007\cdot 46$ (by summing the inequality over all 46 good sets). To obtain a contradiction it suffices to show that there exists a good set with size at least $10$. Consider the set $\{1,6,11,\dots,46\}$. Clearly $a-b+c-d+e\equiv 1\pmod 5$ always, but since $-89\le a-b+c-d+e\le 136$ it follows that if $47\mid a-b+c-d+e$ then $a-b+c-d+e\in \{-47,0,47,94\}$ which is a contradiction. $\blacksquare$

For $k=1,2,\ldots,46$, let $X_k$ be the set of all elements of $X$ congruent to one of $19k,20k,\ldots,28k$. The expected value of $|X_k|$ is $10000 \cdot \tfrac{10}{46}>2007$, so there exists a value of $k$ with $|X_k|>2007$. Let $Y$ be a $2007$-element subset of $X_k$. Then, for all $a,b,c,d,e \in Y$, $a-b+c-d+e$ can take all multiples of $k$ from $(3 \cdot 19-2 \cdot 28)k=k$ to $(3 \cdot 28-2 \cdot 19)k=46k$, but $ak \not\equiv 0 \pmod{47}$ for $1 \le a \le 46$, so $47 \nmid a-b+c-d+e$ and we are done.

We work with a $\mathbb{F}_{47}$ multiset. For $1 \leq i \leq 46$ let $x_i$ denote the number of elements of $X$ congruent to $i \pmod{47}$, so $\sum_{i=1}^{46} x_i=10000$. We will prove that there exists a set $S \subseteq \{1,\ldots,46\}$ such that $0 \not \in S-S+S-S+S$ and $\sum_{i \in S} x_i\geq 2007$. We will consider sets of the form $S_d=\{19d,20d,\ldots,28d\}$ for $d \neq 0$. We can verify that $S_d-S_d+S_d-S_d+S_d=\{d,\ldots,46d\}$, so these sets satisfy the first condition. On the other hand, since $kd$ will cover $\{1,\ldots,46\}$ evenly where $k \neq 0$ and $d$ ranges from $1$ to $46$, so $S_d$ across all $d$ evenly cover $\{1,\ldots,46\}$ as well. Thus by expectation, we expect some $S_d$ to satisfy $$\sum_{i \in S_d} x_i \geq \frac{10000\cdot 10}{46}\geq 2007,$$as desired. $\blacksquare$ Remark: The motivation for this solution primarily contains two parts. First, after using the multiset idea, we would expect that we would have to find $|S|\geq 10$, because otherwise we would not expect $\sum_{i \in S} x_i\geq 2007$ on arbitrary $X$. On the other hand, by Cauchy-Davenport we have $|S-S+S-S+S| \geq \max\{47,5|S|-4\}$, so we should almost certainly have $|S|=10$ and have equality everywhere in the Cauchy-Davenport application, so $S$ should be an arithmetic series.

For an integer $n$, let $S_n = \{x \in X \mid x \equiv n \pmod{47}\}$. Then, for each integer $1 \le m \le 46$, let $T_m = S_{19m} \cup S_{20m} \cup S_{21m} \cup \dots S_{28m}$. Call a set $A$ of integers good if there does not exist $a, b, c, d, e \in A$ such that $47 \mid a - b + c - d + e$. We want to find a good set with at least 2007 elements. Claim: $T_m$ is good for all $m$. Proof: It suffices to show that $a + c + e \not \equiv b + d \pmod{47}$ for all $a, b, c, d, e \in T_m$. Furthermore, it suffices to prove the claim for $m = 1$, since the claim for other $m$ follows from multiplying $a + c + e \not \equiv b + d \pmod{47}$ by $m$ on both sides (allowed since 47 is prime and $47 \nmid m$). Assume $m=1$. Then all elements of $T_m$ are congruent to one of $19, 20, 21, \dots, 28$ modulo 47. $b+d$, the sum of two such elements, is congruent to one of $38, 39, \dots, 56$ modulo 47, or in other words, one of $38, 39, \dots, 46, 0, 1, 2, \dots, 9$. Similarly $a+c+e$ is congruent to one of $57, 58, \dots, 84$, or one of $10, 11, \dots, 37$. Therefore $a+c+e \not \equiv b+d \pmod{47}$ as desired. $\square$ Now, \[\sum_{i=1}^{46} |T_i| = \sum_{k=19}^{28}\sum_{i=1}^{46} |S_{ik}| = 10\sum_{i=1}^{46} |S_i| = 10|X| = 100000,\]so the largest $T_i$ must have at least $\frac{100000}{46} > 2007$ elements, as desired.

The key is the following construction of the set $S_k$ made of the elements in X that are $\equiv{19k,20k,...,28k}\forall 1\le k\le 46$, all taken mod 47. Noting that 47 is prime so these multiples for varying k form a complete residue class 10 times, we get that the expected value of the size of $S_i=10000\frac{10}{46}>2007$, so we can take a suitable $S_i$ that has size at least 2007 and take 2007 elements from there. Then it's obvious that this new set satisfies the problem, because it can get values i from $$3\cdot 19k-2\cdot 28k\equiv k\le i\le 3\cdot 28k-2\cdot 19k\equiv 46k\forall i=jk,$$which can never become 0 mod 47 since i remains a multiple of k and j never gets to a multiple of 47; in particular, the property does indeed satisfy the problem. $\blacksquare$

We work entirely $\pmod {47}$ and let $X$ be a multiset of residues from $\{1,2,\dots,46\}$. Let $A_i={-9i,-7i,-5i,-3i,-i,i,3i,5i,7i,9i}$ for $i=1,\dots,46$. Consider $A_1+A_1+A_1$, that is, $\{x+y+z\mid (x,y,z)\in A_1^3\}$. This spans all odd numbers $\{-27,-25,-23,\dots,27\}$. $A_1+A_1$ spans all even numbers $-18,-16,\dots,18$. It is easy to see that these two sets are disjoint. Multiplying by $i$ does not affect disjointness. Listing out all such sets lists each number ten times. Note that if for some $i$, a subset $Y\subset X$ only has elements from $A_i$ then since $A_i+A_i+A_i$ is disjoint from $A_i+A_i$ this set does have any $a,b,c,d,e$ for which $47\mid a-b+c-d+e$. Now, assume that there is no size-$2007$ subset with elements exclusively in one of the $A_i$s. This means that for each $A_i$, its elements are in $X$ at most $2006$ times. Over all $46$ $i$s and dividing by $10$ to correct overcounting, we get that $|X|\le 2006\cdot 4.6 < 10000$ which is a contradiction.

Quite a nice problem. We can actually solve this quite methodically, as we will show below. The idea is that when the distribution of residues is unbalanced, we can just pick one of them and win, which suggests that the "worst case" is when it is balanced. We see that we need to pick $10$ different residues in that case. Furthermore, we will restrict ourselves by picking either all of a certain residue or nothing. Thus, we first construct a set of $10$ residues such that no triple has the same sum as a pair. The idea is to use parity. If we have $$-9, -7, -5, -3, -1, 1, 3, 5, 7, 9$$then the largest value is $9\cdot 5=45$, so the multiple of $47$ must be $0$. However, $0$ is not possible by parity, so this construction works. We are almost done. This construction works in the "average case". This suggests that a "global" method considering similar constructions should work. Indeed, consider the $46$ sets of $10$ residues formed by multiplying the above construction by an integer from $1$ to $46$. Since they are all relatively prime to $47$, each of these $46$ constructions still work. However, considering all of these together, they cover every residue equally. Thus, at least one of them has at least $$10000\cdot \frac{10}{46}>2007,$$so we are done.

This is a combinatorics problem in disguise :/ Consider the sets given by $A_i = \{-9i, -7i, \dots, 7i, 9i\}$ for $1 \leq i \leq 46$. Every nonzero residue appears in exactly $10$ of these sets. Thus, considering the $100000$ distinct pairs $(j, A_i)$ where $j \in X$ falls in set $A_i$, some $\frac{10^5}{46} > 2007$ of these pairs are present in the same set $A_i$. Picking $Y$ to be these $2007$ elements works. Remark: I spent a ridiculously long amount of time explicitly partitioning $X$ into sets that satisfied the condition (because like, this is not supposed to be a combinatorics problem, right?). The best construction I was able to give was the following: \begin{align*} A_1 &= \{-9, -7, -5, -3, -1, 1, 3, 5, 7, 9\} \\ A_2 &= \{19, 20, 21, 22, 23, 24, 25, 26, 27, 28\} \\ A_3 &= \{10, 11, 12, 13, 14, 33, 34, 35, 36\} \\ A_4 &= \{2, 6, 18, 31, 37, 41, 45\} \\ A_5 &= \{4, 8, 15, 16, 29, 30, 32, 39, 43\}. \end{align*}Unfortunately $A_5$ fails because $4+4+4=4+8$. I would be interested if such an explicit partition existed; then a solution would arise naturally from the fact that one of the $A_i$ has to be a subset of $X$.