The answer is $m = 2n-1$. Observe that if there are two rows, each with at least two marked points, then such quadrilateral must exist. If $m=2n$, consider excluding the row with most marked points, we still have at least $2n-n=n$ marked points in the rest $n-1$ rows. From pigeonhole, there must be a row with at least two marked points. Hence, there are two rows, each with at least two marked points. On the other hand, if $m=2n-1$, then we can mark $(1,k)$ and $(k,1)$ for every $k=1,2,…,n$.