This problem is related to the problem from last week. Last week, we considered an 8 by 8 grid where two diagonally opposite corner squares are removed. It is impossible to cover the 62 remaining squares with rectangles size 2 by 1.
There is an elegant way to prove this. If we think of the grid as a checkerboard, then the two corners will be the same color. Every rectangle will cover a black and a white square. When you have covered 60 squares with 30 rectangles, you will be left with two squares of the same color, which cannot be adjacent.
This week: Take an 8 by 8 checkerboard and remove two squares of opposite color. It is always possible to cover the remaining 62 squares with 2 by 1 rectangles. Make the argument why this is always possible.