4. What is the minimum number of cells that need to be marked in a $50 \times 50$ table so that each vertical or horizontal strip of $1 \times 6$ contains at least one marked cell.
Answer: 416.

Solution. A $50 \times 50$ square can easily be cut into four rectangles of $24 \times 26$ and a central square of $2 \times 2$. Each rectangle can be cut into $4 \cdot 26=104$ strips of $1 \times 6$. Each such strip must have its own marked cell, so there will be no fewer than 416 such cells.

We will show how to mark 416 cells in the required manner. Mark all parallel diagonals with lengths of $5, 11, 17, 23, 29, 35, 41$, and $47$. In total, there will be $2 \cdot (5+11+17+23+29+35+41+47)=416$ marked cells.