$a_i=1$ is a counterexample.

vwu wrote: $a_i=1$ is a counterexample. The problem is fixed!

Here I am assuming that the sequence is strictly increasing (actually unbounded is enough for what follows). Take $p>2020$ to be a prime number. We may assume, maybe by considering the sequence after some point, that $a_1 > p^2$. Also, assume for cleanness that $n$ can be $0$. By looking at the partial sums $$a_1, a_1+a_2 ,... , a_1 + a_2 +...+ a_p$$ we get two cases: First Case: $p|a_1+a_2+...+a_p$. In this case we just take $n=0, k=p, m= \frac{a_1 + a_2 +...+ a_p}{p} - p >0$ and get a contradiction. Second Case: $p|a_i + a_{i+1} +...+a_{i+l}$ for some $1 \le i \le i+l \le p$ and $l<p-1$. Just take $n=i-1 , k=l+1, m= p-k>0$ and get a contradiction. By the PGP we follow in one of the two cases above and we are done.

BUMP!!! The above is incorrect since m is positive and cannot take $\frac{a_1 + a_2 +...+ a_p}{p}$