Consecutive sum divisible by the sequence length ($\exists 0\le i<j\le n:n\mid {\sum }_{k=i+1}^j{a}_k$)

For any sequence of $n\ge 1$ integers, some nonempty consecutive block has sum divisible by $n$. Form the $n+1$ prefix sums $s_0=0,s_j={\sum }_{k=1}^j{a}_k$. There are only $n$ residue classes modulo $n$, so the pigeonhole principle gives $s_i\equiv s_j(\mathrm{mod}n)$ 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.