Let $k$ and $n$ be integers such that $k \geq 2$ and $k \leq n \leq 2 k-1$. Place rectangular tiles, each of size $1 \times k$ or $k \times 1$, on an $n \times n$ chessboard so that each tile covers exactly $k$ cells, and no two tiles overlap. Do this until
no further tile can be placed in this way. For each such $k$ and $n$, determine the minimum number of tiles that such an arrangement may contain.
The required minimum is $n$ if $n=k$, and it is $\min (n, 2 n-2 k+2)$ if $k < n$. There exists a maximal arrangement of cardinality $2 n-r-c$.

Use again the board $[0, n] \times[0, n]$ to stack $k-r$ horizontal tiles in each of the rectangles $[0, k] \times[0, k-r]$ and $[k-c, 2 k-c] \times[k, 2 k-r]$, erect $k-c$ vertical tiles in each of the rectangles $[0, k-c] \times[k-r, 2 k-r]$ and $[k, 2 k-c] \times[0, k]$, then stack $n-2 k+r$ horizontal tiles in the rectangle $[k-c, 2 k-c] \times[2 k-r, n]$, and erect $n-2 k+c$ vertical tiles in the rectangle $[2 k-c, n] \times[1, k+1]$.