Let $A$ be the $n\times n$ matrix of all $1$s. Let $I$ be the $m\times m$ identity matrix with $m<n$. Prove that any $n\times n$ block of $A\otimes I$ contains a $1$.
My approach is to show that the largest possible block of all zeros in $A\otimes I$ is $k\times k$ where $k=floor(m/2)$. If true, then no $n\times n$ block could contain all zeroes.
For convenience, Let $B = A \otimes I_m$. Then, $B$ is an $mn \times mn$ block matrix where each block is an $m \times m$ identity matrix $I_m$, that is $$ B = \begin{pmatrix} I_m & \ldots & I_m \\ \vdots & \ddots & \vdots \\ I_m & \ldots & I_m \\ \end{pmatrix}. $$ Then, $B$ can be written element-wise as $$ B(i,j) = \begin{cases} 1 ,& \text{if } (i-j) \text{ is divisible by } m\\ 0, & \text{otherwise} \end{cases} ~~~~~ \text{ for } i, j = 1, \ldots, mn. $$
It is trivial to show that, for a fixed $i'$th row of matrix $B$, $$ B(i',j) = 1 ~~~~~ \text{ for } j = i', i' + m, \ldots,i' + m (n-1) $$ and the other elements on row $i'$ are zero. Then, there is a $1$ in every consecutive $m$ elements. It follows that, there is a $1$ in every consecutive $n$ elements since $n > m$. Hence, any $n \times n$ block of $B$ contains a 1.