Mongolian Mathematical Olympiad

Assume that $a_0,a_1,\dots ,a_{k-1}$ is a sequence of integers satisfying the given condition, where we take indices modulo $k$. If each term of the sequence is divisible by $m$ then the sequence $a_0/m,a_1/m,\dots a_{k-1}/m$ satisfies the given condition. Hence we can assume that $a_0$ is not divisible by $m$.

We claim that the sequence contains at least $m$ consecutive $1$ if $(k,m)=1$. Let $S_i=a_i+\cdots +a_{i+m-1}$ $(0\le i\le k-1)$ and let $t$ be an index such that $S_t$ is the smallest power of $m$. Clearly, $S_i$ is divisible by $S_t$ for each index $i$ and therefore $a_{i+m}-a_i=S_{i+1}-S_i\equiv 0(\mathrm{mod}{S}_t)$. Fix integers $x,y$ such that $xk+ym=1$. Then $a_{i+1}\equiv a_{i+xk+ym}\equiv a_i(\mathrm{mod}{S}_t)$ which implies that $a_0\equiv a_1\equiv \cdots \equiv a_{k-1}(\mathrm{mod}{S}_t)$. Since $m{a}_0\equiv m{a}_t=S_t\equiv 0(\mathrm{mod}{S}_t)$ and $m\nmid a_0$ we get $S_t=m$ and so $a_t=a_{t+1}=\cdots =a_{t+m-1}=1$. The claim is proved.

Now assume that we have a sequence of $mn-1$ integers satisfying the given condition and $a_0$ is not divisible by $m$. The sequence contains $m$ consecutive $1$ by the claim. Delete one of them and then the remaining sequence of $mn-2$ integers satisfies the given condition. We have $(mn-2,k)=1$ since $n,m$ are odd. So the remaining sequence contains at least $m$ consecutive $1$ by the claim, completing the solution.