19. In an $m \times n$ rectangular table filled with natural numbers, you can add an integer $k$ to two adjacent cells simultaneously, ensuring the resulting numbers are non-negative integers (two cells sharing a common edge are called adjacent). Determine the necessary and sufficient condition so that after a finite number of such operations, all numbers in the table can be 0.
19. In an $m \times n$ table, adjacent cells are colored with two different colors * and $\square$, and the sums of the numbers in the two types of cells are denoted as $S_{*}$ and $S_{\square}$, respectively. Let $S = S_{*} - S_{\square}$. Since $S$ remains unchanged after each operation, $S = 0$ is a necessary condition for all numbers in the table to become 0 after several operations.

Now, we prove that $S = 0$ is also a sufficient condition. Starting from the first column, let the number in the first row of the first column be $a$, the number in the second row of the first column be $b$, and the number in the third row of the first column be $c$.

If $a > b$, add $a - b$ to both $b$ and $c$, and then add $-a$ to both $a$ and $b + (a - b) = a$.
If $a \leq b$, add $-a$ to both $a$ and $b$.
Continue this process until the table becomes:
\begin{tabular}{|c|c}
\hline 0 & \\
0 & \\
$\vdots$ & \\
0 & $r$ \\
$g$ & $h$ \\
\hline
\end{tabular}
$$
\left\lvert\, \begin{array}{|l|l|}
\hline 0 \\
\vdots \\
0 & \\
0 \\
& h \\
0
\end{array}\right.
$$

If $g \leq h$, add $-g$ to both $g$ and $h$. If $g > h$, add $g - h$ to both $r$ and $h$, and then add $-g$ to both $g$ and $h + (g - h)$. In any case, we can make all the numbers in the first column 0. Continuing this process, we can make all the numbers in the table 0 except possibly one number in the $n$-th column.
Since $S = 0$, at this point, every number must be 0.