Answer: $n+1$ Proof: By induction. Let $A_n$ be the set $A$ when $S=\{1,2,...,n\}$. Consider a post covering $(n,n)$. - If the post is at $(n,n)$, then all points in $A_{n-1}$ aren't covered by that post. By induction, $n$ posts are required to cover $A_{n-1}$. - If the post is at $(n,n-1)$, then the point $(n-1,n)$ isn't covered by that post, so it needs another post to cover it. Observe that all points in $A_{n-2}$ aren't covered by these two posts. By induction, $n-1$ posts are required to cover $A_{n-2}$. - If the post is at $(n-1,n)$, then it's symmetric to the second case.