This problem is found in Richard A. Brauldi's book on Introductory Combinatorics. It goes as follows:
Given m integers $a_1, a_2, ... ,a_m$, there exist integers $k$ and $l$ with $0 \le k \lt l \le m$ such that $a_{k+1} + a_{k+2} + ... + a_l$ is divisible by $m$. Less formally, there exist consecutive $a$'s in the sequence $a_1, a_2, . .. ,a_m$ whose sum is divisible by $m$.
The author starts the proof like this:
Consider the $m$ sums:
$a_1, a_1 + a_2, a_1 + a_2 + a_3, ..., a_1 + a_2 + a_3 + ... + a_m$
If any of these sums are divisible by $m$, then the conclusion holds. Thus, we may suppose that each of these sums has a nonzero remainder when divided by $m$ and so ...
The author then continues with the proof; yet, I haven't understood how it is that we can just suppose that the sums have a nonzero remainder? I at first thought the author was attempting at a proof by contradiction, but the author continues to demonstrate how it is possible to construct with integers $k$ and $l$ the sequence $a_{k + 1}, ..., a_l$, the sum of the terms of which is divisible by $m$. But this doesn't contradict the supposition we've took previously, because we've only considered series beggining with the first term, $a_1$ and not any general sequence. Could anyone elucidate the flaw in my reasoning? That you in advance.
Because of this, it suffices to prove that the conclusion holds when none of those sums is divisible by $m$, i.e. the remainders are nonzero.