59th Ukrainian National Mathematical Olympiad

We will start with counting polyominos of smallest area. There is only 1 polyomino of area 1, 1 polyomino of area 2, and 2 polyominos of area 3. Thus, the smallest area of a rectangle that consists of 8 polyominos is $1\cdot 1+1\cdot 2+2\cdot 3+4\cdot 4=25$. But then it has to be of a size $25\times 1$ (since $5\times 5$ square doesn't satisfy the condition). It is clear for such a rectangle that it can only be split into rectangles of size $k\times 1$ and, but such a rectangle has an area at least $1+2+...+8=36>25$. Thus, the smallest possible area is 26. An example of splitting a $13\times 2$ rectangle is shown on Fig. 18.