TITLE: Let $a_1, \ldots,\ a_{100}$ be $100$ positive integers. Show that for some $m,\ n$ with $1\le m\le n\le 100, \sum_{i=m}^n a_i$ is divisible by $100$.
QUESTION [6 upvotes]: Let $a_1,\ a_2,\ a_3, \ldots,\ a_{100}$ be $100$ positive integers. Show that for some $m,\ n$ with $1\le m\le n\le 100, \sum_{i=m}^n a_i$ is divisible by $100$.

REPLY [13 votes]: Consider the partial sums $s_n=\sum_{i=1}^n a_i$, where $1\leq n\leq100$. If $s_i$ is divisible by $100$ for some $i$ then we are done. Otherwise, since there are $100$ of the $s_i$ and $99$ possible remainders upon division by $100$, two of the $s_i$, say $s_n$ and $s_m$ with $m>n$, must be equal modulo $100$. Then $s_n-s_m = \sum_{i=m}^n a_i$ is divisible by $100$, as desired.