Solution:

We shall show that the desired value is equal to $4$.

Let $(i, j)$ be the unit square in the $i$-th row and the $j$-th column of a $4 \times 4$ square. It is easy to check that if the squares $(1,1), (1,2), (2,1), (2,3), (3,2), (3,4), (4,3), (4,4)$ are white and the others are black, then the condition of the problem is satisfied.

To prove that for $n \geq 5$ there is no coloring with the required properties, it is enough to deal only with the case $n=5$ (any $n \times n$ square contains a $5 \times 5$ square). Consider an arbitrary black-white coloring of a $5 \times 5$ square. At least $13$ of the unit squares have the same coloring, for example, black. Three cases are possible.

Case 1. One of the rows, say $l$, contains only black squares. Then one of the other rows contains at least $2$ black squares. Hence the corners of the rectangle with two vertices at these squares and the other two in $l$ are black.

Case 2. One of the rows, say $l$, contains exactly $4$ black squares. Then one of the other rows contains at least $3$ black squares. Then at least two of them do not correspond to the white square in $l$. Hence the corners of the rectangle with two vertices at these squares and the other two in $l$ are black.

Case 3. Any row contains at most $3$ black squares. Then at least $3$ rows contain exactly $3$ black squares, say the rows $1,2,3$ from above. We call a column black (white) if its upper square is black (white). There are exactly $3$ black columns. If a row contains $2$ black squares that lie in black columns, then obviously there is a rectangle with black corners. Otherwise, two of the black squares in row $2$ and two of the black squares in the two white columns are black corners of a rectangle.

The solution is completed.