Monochromatic Rectangle Tiling
As can be seen in the comments, Jukka provides a $2 \times 4$ counter example, which is further generalized by Peter to $4 \times 4$ to meet the requirements. Both analysis and program verification (use dynamic programming to calculate $\chi'$) can be performed to make sure $\chi \neq \chi&#39;$ is possible when $n = 4$.

Actually Peter's generalization can be used to show there is an $n \times n$ matrix s.t. $\chi \neq \chi&#39;$ for any $n$. The counter example is shown below:

$\begin{bmatrix}
1 1 0 0 0 \cdots 0 \\
0 1 1 0 0 \cdots 0 \\
0 0 0 0 0 \cdots 0 \\
0 0 0 0 0 \cdots 0 \\
\vdots \\
0 0 0 0 0 \cdots 0
\end{bmatrix}$

There's a unique column with two 1's. At least one continuous rectangle is needed to cover the 0's in this column. And this continuous rectangle doesn't cover any elements in the two rows corresponding to the two 1's. Remove this continuous rectangle and the rest will be a $2 \times n$ matrix:

$\begin{bmatrix}
1 1 0 0 0 \cdots 0 \\
0 1 1 0 0 \cdots 0
\end{bmatrix}$

which needs at least 5 continuous rectangles to cover. So $\chi&#39; \geq 6$. However, it is easy to see $\chi \leq 5$. So $\chi \neq \chi&#39;$.