TITLE: A Pigeonhole-Principle from IMO Shortlist.
QUESTION [13 upvotes]: I have solved a great many pigeonhole principle problems so far.
Now I am stuck with this.The statement of the problem runs as follows-

Let $p$ be an odd prime. Consider $p-1$ positive integers $x_1,x_2,x_3,...,x_{p-1}$, none of which is divisible by $p$. Then consider all possible sums of the form $\pm x_1 \pm x_2 \pm x_3 \pm \cdots \pm x_{p-1}$. Clearly there are $2^{p-1}$ such sums. Prove that at least one among these sums is divisible by $p$.

REPLY [7 votes]: How about using the Cauchy-Davenport theorem?
Namely, consider the sets $A_i = \{x_i, -x_i\}$. Since $p$ is odd, $|A_i|=2$ for all $i$. Now observe that 
$$ |A_1 + A_2| \geq 2 + 2 - 1 = \min\{p, 3\}$$
One can now show by induction that
$$|A_1 + \ldots + A_k| \geq \min\{p, k+1\}$$
Indeed, assuming the hypothesis, the theorem gives us
$$|A_1 + \ldots + A_k + A_{k+1}| \geq \min\{p, \min\{p, k+1\} + 2 - 1\}=\min\{p, k+2\}$$
and thus at the end
$$|A_1 + \ldots + A_{p-1}|\geq \min\{p,p\}=p$$
so in fact we proved something stronger: we can write any residue modulo $p$ as a sum of the form $\pm x_1 \pm\ldots\pm x_{p-1}$.