, 2007

Call a set $M$ of integers good if $47 \nmid a-b+c-d+e$ for any $a, b, c, d, e \in M$.

Consider the set $J=\{-9,-7,-5,-3,-1,1,3,5,7,9\}$. We claim that $J$ is good. Actually, for any $a, b, c, d, e \in J$ the number $a-b+c-d+e$ is odd and
$$-45 = (-9) - 9 + (-9) - 9 + (-9) \leq a-b+c-d+e \leq 9 - (-9) + 9 - (-9) + 9 = 45.$$
But there is no odd number divisible by $47$ between $-45$ and $45$.

For any $k = 1, \ldots, 46$ consider the set
$$A_{k} = \{x \in X \mid \exists j \in J: \quad kx \equiv j \pmod{47}\}.$$
If $A_{k}$ is not good, then $47 \mid a-b+c-d+e$ for some $a, b, c, d, e \in A_{k}$, hence $47 \mid k a - k b + k c - k d + k e$. But set $J$ contains numbers with the same residues modulo $47$, so $J$ also is not good. This is a contradiction; therefore each $A_{k}$ is a good subset of $X$.

Then it suffices to prove that there exists a number $k$ such that $|A_{k}| \geq 2007$. Note that each $x \in X$ is contained in exactly $10$ sets $A_{k}$. Then
$$\sum_{k=1}^{46} |A_{k}| = 10|X| = 100000$$
hence for some value of $k$ we have
$$|A_{k}| \geq \frac{100000}{46} > 2173 > 2007$$