Solution:

Answer: 16

Suppose Mike writes $k$ $M$'s. Let $a$ be the number of squares which, if Harry writes an $H$ in, will yield either $H M M$ or $M M H$ horizontally, and let $b$ be the number of squares which, if Harry writes an $H$ in, will yield either $H M M$ or $M M H$ vertically. We will show that $a \leq k$ and $b \leq k$. Then, it will follow that there are at most $a+b \leq 2k$ squares which Harry cannot write an $H$ in. There will be at least $64-k-2k=64-3k$ squares which Harry can write in. If $64-3k \geq k+1$, or $k \leq 15$, then Harry wins.

We will show that $a \leq k$ (that $b \leq k$ will follow by symmetry). Suppose there are $a_{i}$ $M$'s in row $i$. In each group of 2 or more consecutive $M$'s, Harry cannot write $H$ to the left or right of that group, giving at most 2 forbidden squares. Hence $a_{i}$ is at most the number of $M$'s in row $i$. Summing over the rows gives the desired result.

Mike can win by writing 16 $M$'s according to the following diagram: