Coloring a $9\times9$ Board Consider a $9 \times 9$ board, and we want to paint each of the $81$ squares either white or black. There are $64$ $2\times2$ squares within the $9\times9$ board, and there are $16$ different $2\times2$ "paintings" (for example, $\begin{pmatrix} W&B\\B&W\end{pmatrix}$, $\begin{pmatrix} B&B\\B&W\end{pmatrix}$, etc.).
Question: By Pigeonhole Principle, at least one of the $16$ $2\times2$ paintings must appear at least $\frac{64}{16} = 4$ times. Is there, however, a realization that uses every $2\times2$ painting exactly $4$ times?
Best Attempt: I have the following $9\times9$ masterpiece, which uses $14$ $2\times2$ paintings exactly 4 times but uses $\begin{pmatrix} B&B\\B&W\end{pmatrix}$ $5$ times and $\begin{pmatrix} W&B\\B&B\end{pmatrix}$ only $3$ times. See here:

 A: Tile the plane with copies of the $4\times4$ pattern
$$\pmatrix{0&0&0&1\\0&0&1&0\\1&0&1&1\\0&1&1&1}$$
(using, say $0$ for Black and $1$ for White).  Then any $9\times9$ subarray will contain the $16$ different $2\times2$ patterns exactly $4$ times.
The essential idea here is to identify opposite edges of the $4\times4$ array, making a torus out of the square.  It's a bit laborious, but not all that hard, to check that the $16$ different $2\times2$ patterns occur exactly once.  Four copies of it, therefore, again with toroidal identification of edges, contain the $16$ patterns $4$ times each.  Buffering four copies with an extra row and column simply makes the toroidal identification explicit.
Remark:  Peter's $9\times9$ solution is not of this "toroidal-based" form.  
A: [0 1 0 0 1 1 0 0 0]

[1 1 1 0 1 0 1 1 1]

[1 1 0 1 0 0 0 0 1]

[1 1 1 1 0 0 1 0 0]

[0 1 0 1 1 1 1 1 1]

[1 0 1 0 1 1 1 0 0]

[0 0 1 0 0 0 0 0 1]

[1 0 1 0 1 0 0 0 1]

[0 0 0 1 1 0 0 1 0]

should be a coloring satisfying the condition! Can someone verify it ?
