From an olympic book:
Let $m$ be an odd number and $a_1, ... , a_m$ integers. $x=(x_1,...,x_m)$ is a permutation of the integers $1, ..., m$ and $f$ at $x$ is defined by $f(x)=x_1a_1 + ... + x_m a_m$. Prove that there exist two permutations $x$ and $y$ such that $f(x)-f(y)$ is divisible by $m!$.
I'm aware of a few facts. For instance, if all $a_i$ are odd (or even), any permutations of $x$ results in values of $f$ of the same parity, but there are only $m!/2$ (for $m>1$) residues module $m!$ of the same parity, and an application of the Pigeon Hole Principle yields the conclusion. So in the general case I'd like to say something like there's a greater number of permutations that give a value of $f$ of one parity than the other. I don't know how to show this.