Answer: $2n-1$. Example: For $n=3$ we have $\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|} \hline 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 1 & 1 & 0 & 0 \\ \hline 0 & 1 & 0 & 0 & 1 & 0 \\ \hline 0 & 1 & 0 & 0 & 1 & 0 \\ \hline 0 & 0 & 1 & 1 & 0 & 0 \\ \hline 0 & 0 & 0 & 0 & 0 & 0 \\ \hline \end{tabular}$ For $n\geq 5,$ we use induction.

Proof: Divide the table into $2\times2$ parts. Let there be $f(i)$ times $2\times 2$ squares which contain $i$ black squares. \[2(n-1)^2=f(1)+2f(2)+3f(3)+4f(4)\geq f(1)+2(f(2)+f(3)+f(4))=f(1)+2(n^2-f(0)-f(1))=2n^2-2f(0)-f(1)\]\[2f(0)+2f(1)\geq 2f(0)+f(1)\geq 4n-2\implies f(0)+f(1)\geq 2n-1\]Thus there must exist at least $2n-1$ times $2\times 2$ squares which can contain $3-$square corner as desired.$\blacksquare$

Similar to @ above but posting for storage The answer is $\boxed{2n-1}$. Construction: We will proceed by induction. For $n=3$ the following construction can be verified to work [asy][asy] filldraw(((1,2)--(1,4)--(2,4)--(2,2)--cycle), gray); filldraw(((4,2)--(4,4)--(5,4)--(5,2)--cycle), gray); filldraw(((2,1)--(4,1)--(4,2)--(2,2)--cycle), gray); filldraw(((2,4)--(4,4)--(4,5)--(2,5)--cycle), gray); add(grid(6,6)); [/asy][/asy] When going from $n\mapsto n+1$ place a copy for the $2n\times 2n$ construction for $n$ in the center of the $2n+2\times 2n+2$ grid. Then with the additional $2(n+1)^2-2(n-1)^2=4n$ squares surround the central $2n\times 2n$ construction. Then you are able to cut out $4$ additional three corner pieces from the corners in addition to the $2n-1$ three corners from the center. Bound: Divide the grid into $2\times 2$ squares. In order to stop a three corner from being cut out from one of these squares there must be at least two squares shaded. Notice there can only by $(n-1)^2$ such squares with at least two squares shaded. Thus we can cut out three corners form at least $n^2-(n-1)^2=2n-1$ of them.

What is three-square corner?

geoispowerful wrote: What is three-square corner? Square $2 \times 2$ without one (any) cell.