A natural number is written in each square of an $ m \times n$ chess board. The allowed move is to add an integer $ k$ to each of two adjacent numbers in such a way that non-negative numbers are obtained. (Two squares are adjacent if they have a common side.) Find a necessary and sufficient condition for it to be possible for all the numbers to be zero after finitely many operations.
1. **Define the Invariant:**
   Let \( S_b \) and \( S_w \) be the sums of the numbers on the black and white squares of the \( m \times n \) chessboard, respectively. Note that the chessboard is colored in a checkerboard pattern, so each square is either black or white.

2. **Invariant Property:**
   The difference \( S_b - S_w \) is an invariant under the allowed operations. This is because adding an integer \( k \) to two adjacent squares (one black and one white) will increase both \( S_b \) and \( S_w \) by \( k \), leaving \( S_b - S_w \) unchanged.

3. **Final Condition:**
   At the end, we want all numbers to be zero. This implies \( S_b = 0 \) and \( S_w = 0 \). Therefore, \( S_b - S_w = 0 \) must hold at the beginning as well. Hence, a necessary condition is \( S_b = S_w \) initially.

4. **Sufficiency of the Condition:**
   We need to show that if \( S_b = S_w \) initially, it is possible to make all numbers zero using the allowed operations.

5. **Algorithm to Zero Out the Numbers:**
   Consider three consecutive cells \( a_1, a_2, a_3 \) where \( a_1 \) is adjacent to \( a_2 \) and \( a_2 \) is adjacent to \( a_3 \). We will handle two cases based on the relative values of \( a_1 \) and \( a_2 \):

   - **Case 1:** \( a_1 \geq a_2 \)
     - Add \( a_1 - a_2 \) to both \( a_2 \) and \( a_3 \). This operation makes \( a_2 \) equal to \( a_1 \).

   - **Case 2:** \( a_1 \leq a_2 \)
     - Add \( -a_1 \) to both \( a_1 \) and \( a_2 \). This operation makes \( a_1 \) equal to zero.

6. **Applying the Algorithm:**
   Apply the above operations row by row. After processing each row, the last two entries in the row will be equal because \( S_b = S_w \). This ensures that the sum of the numbers in each row can be reduced to zero.

7. **Conclusion:**
   Since we have shown that the condition \( S_b = S_w \) is both necessary and sufficient for it to be possible to make all numbers zero, we conclude the proof.

\(\blacksquare\)

The final answer is \( \boxed{ S_b = S_w } \)