= Consecutive sum divisible by the sequence length
{title2=$\exists\,0\leq i<j\leq n:\ n\mid\sum_{k=i+1}^{j}a_k$}
For any <sequence> of $n\geq1$ <integers>, some nonempty consecutive block has sum divisible by $n$. Form the $n+1$ <prefix sums> $s_0=0,s_j=\sum_{k=1}^ja_k$. There are only $n$ <residue classes> modulo $n$, so the <pigeonhole principle> gives $s_i\equiv s_j\pmod n$ for some $i<j$. <Subtraction> proves that $s_j-s_i$ is the required block sum. Negative entries and repeated entries cause no difficulty. Applied after any ordering of a <finite set>, it also gives a nonempty <subset> whose sum is divisible by the size of the <set>.
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