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 $S_i = a_i + \dots + a_{i+m-1}$ $(0 \le i \le k-1)$ 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 $a_{i+m} - a_i = S_{i+1} - S_i \equiv 0 \pmod{S_t}$. Fix integers $x, y$ such that $xk + ym = 1$. Then $a_{i+1} \equiv a_{i+xk+ym} \equiv a_i \pmod{S_t}$ which implies that $a_0 \equiv a_1 \equiv \dots \equiv a_{k-1} \pmod{S_t}$. Since $ma_0 \equiv ma_t = S_t \equiv 0 \pmod{S_t}$ and $m \nmid a_0$ we get $S_t = m$ and so $a_t = a_{t+1} = \dots = a_{t+m-1} = 1$. 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.