In the sequel, we assume that all of $ab$ boxes of the original grid can be marked by repeating the given procedure a certain number of times after we reach the situation where $N$ of the boxes are marked, and we show that $N \geq (a - 1)(b - 1)$.

Suppose the last marked box to attain the goal of marking all of the $ab$ boxes lies on the $X$-th row and $Y$-th column. Then, we see that the sum of the number of rows and the number of columns on which the markings were performed prior to the last marking and after the marking of $N$ boxes are achieved is at most $a + b - 2$. Furthermore, markings cannot be repeated consecutively on any row or column. Therefore, the number of markings performed after $N$ boxes are marked (including the last marking) is at most $a + b - 1$. Since the number of $\checkmark$ increases by $1$ at each marking, we need, in order to complete the marking of all the $ab$ boxes, to have $N \geq ab - (a + b - 1) = (a - 1)(b - 1)$.

Thus, we conclude that $(a-1)(b-1)$ is the desired answer to the problem.