网格染色与同色矩形

必然存在。

固定一行（例如 $y=0$）。其中必有无限多个点同为黑或同为白。取其中一种颜色（不妨设白色）对应的无限多列集合 $C$。

再看下一行 $y=1$ 在这些列上的颜色：要么在 $C$ 中仍有无限多列为白色（则两行 $y=0,1$ 与任取两列构成同色矩形），要么在 $C$ 中有无限多列为黑色（记为 $C'$）。

继续看 $y=2$ 在 $C'$ 上的颜色：要么有无限多列为黑色（则用 $y=1,2$ 得到黑色矩形），要么有无限多列为白色（则用 $y=0,2$ 得到白色矩形）。

因此一定能找到两行与两列，使四个角同色。

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英文解析

Yes, it must exist.

Fix one row, for example $y=0$. Among the infinitely many lattice points on that row, infinitely many have the same color; call the infinite set of such columns $C$. Now inspect another row, say $y=1$, restricted to the columns in $C$. If infinitely many of those positions have the same color as the first row, any two such columns form a monochromatic rectangle with rows $0$ and $1$. Otherwise, infinitely many positions in row $1$ must have the opposite color; then repeat the same argument with that color. In either case, two rows and two columns can be chosen whose four corner points have the same color.