20 natural numbers between 1 and 1000, find two subsets with the same size and sum 
Let $A=\{a_1,a_2,\ldots,a_{20}\}$ a set of distinct natural numbers between 1 and 1000. Prove that $\exists X,Y\subseteq A$ such that $X\cap Y = \emptyset$, $|X|=|Y|>0$, and $$\sum_{x\in X}x=\sum_{y\in Y}y$$

In my attempt to solve the problem I couldn't find the pigeons and the holes. Even a hint would do. Thanks!

The condition $X \cap Y = \emptyset$ is a red herring.
Suppose you have two $k$-element sets $X$ and $Y$ with the same sum, but they happen to have $\ell>0$ elements in common. Delete those $\ell$ elements, and you have two disjoint $(k-\ell)$-element sets $X \setminus Y$ and $Y \setminus X$, still with the same sum. (The sums both decrease by the sum of the deleted elements.)
Since we now want two sets with the same size and the same sum, it makes sense to look at all the $k$-element subsets of $A$ as the pigeons, with all the possible sums as the holes. (Think about which $k$ makes the most sense to use: you get to pick.)
