A $98 \times 98$ chess board has the squares colored alternately black and white in the usual way. A move consists of selecting a rectangular subset of the squares (with boundary parallel to the sides of the board) and changing their color. What is the smallest number of moves required to make all the squares black?
The solution given is:
There are $4 \cdot 97$ adjacent pairs of squares in the border and each pair has one black and one white square. Each move can fix at most 4 pairs, so we need at least 97 moves. However, we start with two corners one color and two another, so at least one rectangle must include a corner square. But such a rectangle can only fix two pairs, so at least 98 moves are needed.
I have difficulty in understanding the solution specially the bold part. Please can someone help. I understand how 98 can be achieved but cant prove its the minimum, other proofs are welcome as well.