We claim that when $m=n=1$, all pairs of positive integers can appear on the board, otherwise, all pairs of positive integers except $(1, 1)$ can appear on the board. First, we show that $(1, 1)$ can appear on the board iff $m=n=1$. Notice that $(1,1)$ can only appear after $(k,1)$ for some positive integer $k$. But the possible pairs from $(k,1)$ are $(k,k+1),(k+1,k),(k,k)$. So $(1,1)$ is possible only if $k=1$. Notice that we can go from $(a,b) \to (ab,b) \to (ab+b,b) \to (a+1,b)$. Similarly, we can go from $(a,b) \to (a,b+1)$. Now we will show that one can go from any pair $(a,b)$ to $(2,1)$ or $(1,2)$. WLOG assume that $a \ge b$. Then we can go from $(a,b) \to (a,b+1) \to \dots \to (a,2a) \to (a,2) \to (a+1,2) \to \dots \to (2^k, 2) \to (2^{k-1},2) \to \dots \to (2,2)$. Then we can go from $(1,2)$ or $(2,1)$. From that, all pairs of positive integers other than $(1,1)$ can be formed.