#### Solution

This is another “colour the board” problem. This time however, the traditional chessboard pattern is not going to work – each of the rectangular blocks covers two white and two black squares and it doesn’t lead anywhere. $\\$

In order to find a proper colouring, we need to think about what we actually need and want from it. We would like every rectangle to cover the same number of coloured squares. It would be nice if this number were odd, because there is also an odd number of blocks ($25$, like in the first example). We could colour the square in stripes, leaving 3 blank lines between each coloured line – and that would work for vertically aligned rectangles, but what about horizontal ones? We can notice, however, that if we exploit diagonals, we get a desired effect.

In the picture on the left, we can see a colouring such that each rectangle covers exatly one coloured square. In total, there are $24$ coloured squares in the picture. However, to cover the entire board, we would need $25$ rectangles. That is a contradiction, because the rectangles cannot overlap. $\\$

We need to be careful, however, because the diagonal colouring can be done in a couple of different ways and some of them yield $25$ or $26$ coloured squares, which is useless for us as far as this idea is concerned. So if you ever encounter a similar problem, don’t be discouraged when it doesn’t work the first time. $\\$

Another solution is visible on the right. We colour larger $2 \times 2$ squares in chessboard-like pattern. This way, every rectangle covers two coloured squares. However, there are not enough coloured squares! There are only $48$, and if we used $25$ rectangles, as each covers two squares, we would need to cover $50$ in total.