在n个整数里，它们的某些子集和能被$n$整除

Let your set of $n$ integers be $Y = {a_1, a_2, ..., a_n}$. We will prove that there is some nonempty subset $X\subseteq Y$ such that the sum of the elements of $X$ is divisible by $n$.

Let $S_k = a_1+a_2+...+a_k$, for $1\le k\le n$. If any one of $S_1, S_2, ..., S_n$ is divisible by $n$, then we are already done.

If this is not the case, then by the Pigeonhole Principle, there must exist $1\le i\lt j\le n$ such that $S_i\equiv S_j\pmod{n}$. Then, $S_j-S_i\equiv 0\pmod{n}$. But $S_j-S_i=a_{i+1}+a_{i+2}+...+a_j$, meaning that $S_j-S_i$ is also the sum of the elements in some nonempty subset of $Y$. This completes the proof.

It is interesting to note that the method above will give you a subset of $Y$ that has the sum of its elements divisible by $n$, no matter which ordering of the elements of $Y$ you use. This doesn't mean that the statement that we have just proved is weak or trivial, but it does mean that finding a subset that satisfies the desired condition is actually quite easy.