Given a table $m\times n$, in each cell of which a number $+1$ or $-1$ is written. It is known that initially exactly one $-1$ is in the table, all the other numbers being $+1$. During a move, it is allowed to chose any cell containing $-1$, replace this $-1$ by $0$, and simultaneously multiply all the numbers in the neighbouring cells by $-1$ (we say that two cells are neighbouring if they have a common side). Find all $(m,n)$ for which using such moves one can obtain the table containing zeros only, regardless of the cell in which the initial $-1$ stands.
To solve this problem, we need to determine the conditions under which we can transform a table of size \(m \times n\) containing exactly one \(-1\) and all other entries as \(+1\) into a table containing only zeros using the given move. The move allows us to choose any cell containing \(-1\), replace it with \(0\), and simultaneously multiply all the numbers in the neighboring cells by \(-1\).

1. **Initial Setup and Move Description**:
   - Initially, the table has one \(-1\) and all other entries are \(+1\).
   - A move consists of selecting a cell with \(-1\), changing it to \(0\), and flipping the signs of all its neighboring cells (cells sharing a common side).

2. **Effect of a Move**:
   - When a \(-1\) is changed to \(0\), the neighboring cells' signs are flipped.
   - This means that if a neighboring cell was \(+1\), it becomes \(-1\), and if it was \(-1\), it becomes \(+1\).

3. **Parity Consideration**:
   - We need to consider the parity (odd or even nature) of the number of \(-1\)s in the table.
   - Initially, there is exactly one \(-1\), which is odd.
   - Each move changes the number of \(-1\)s by an even number (since flipping the signs of an even number of cells changes the parity of the number of \(-1\)s).

4. **Final Configuration**:
   - To achieve a table of all zeros, the number of \(-1\)s must be zero, which is even.
   - Therefore, the number of \(-1\)s must change from odd to even, which is only possible if the total number of cells in the table is even.

5. **Condition for \(m\) and \(n\)**:
   - The total number of cells in the table is \(m \times n\).
   - For the number of \(-1\)s to change from odd to even, \(m \times n\) must be even.
   - This implies that either \(m\) or \(n\) must be even (or both).

6. **Double Counting Argument**:
   - Consider the parity of the number of \(-1\)s in the table after each move.
   - If \(m\) and \(n\) are both odd, then \((m-1)(n-1)\) is even, and the parity of the number of \(-1\)s can be changed to even.
   - If either \(m\) or \(n\) is even, then \((m-1)(n-1)\) is odd, and the parity of the number of \(-1\)s cannot be changed to even.

Therefore, the condition for transforming the table to all zeros is that \(2 \mid (m-1)(n-1)\).

The final answer is \( \boxed{ 2 \mid (m-1)(n-1) } \).