Proof: Assume it is possible to remove all the cards. Each time an operation is performed, record the number of cards that have already been removed and are adjacent to the card being removed, and let the sum of these numbers be $\sigma$.

The number recorded when the first card is removed is 0. Then, for each card removed, it starts as white, and it turns black only when an odd number of adjacent cards have been removed. Thus, the number recorded is an odd number. Therefore, $\sigma$ is the sum of $m n-1$ odd numbers, hence
$$\sigma=m n-1(\bmod 2) \text {. }$$

On the other hand, consider each pair of adjacent cards. The number of cards not being the one removed is 1. Therefore, $\sigma$ equals the number of all adjacent pairs of small squares on the rectangular board, i.e.,
$$\sigma=m(n-1)+n(m-1) \text {. }$$

To remove all the cards, it must be true that
$$m n-1 \equiv m(n-1)+n(m-1)(\bmod 2) \text {, }$$

which is equivalent to $(m-1)(n-1)$ being even. Thus, the necessary condition for removing all the cards is that at least one of $m$ and $n$ is odd.

Next, we prove that this condition is also sufficient. By symmetry, assume $m=2 k-1$. Let the position of the card be $(i, j)$, where $1 \leqslant i \leqslant m, 1 \leqslant j \leqslant n$, and assume the card at $(1,1)$ is black side up. We remove the cards at $(i, 1)$ in order, $i=1,2, \cdots, m$. If $n=1$, the conclusion holds. Assume $n \geqslant 2$, then all the cards at $(i, 2)$ are black side up. We remove the cards at $(2 i-1,2)$ in order, $i=1,2, \cdots, k$, each card at $(2 i, 2)$ is flipped twice, so it remains black side up, and then we remove these cards. If $n=2$, the conclusion holds. Otherwise, all the cards at $(i, 3)$ are black side up. Repeat the above process until the given $n$.

In summary, the pairs of positive integers $(m, n)$ that allow all the cards to be removed from the rectangular board satisfy: at least one of $m$ and $n$ is odd.