In an $m\times n$ table there is a policeman in cell $(1,1)$, and there is a thief in cell $(i,j)$. A move is going from a cell to a neighbor (each cell has at most four neighbors). Thief makes the first move, then the policeman moves and ... For which $(i,j)$ the policeman can catch the thief?
1. **Define the problem in terms of parity:**
   - The policeman starts at cell \((1,1)\).
   - The thief starts at cell \((i,j)\).
   - We need to determine for which \((i,j)\) the policeman can catch the thief.

2. **Understand the movement and distance:**
   - Both the policeman and the thief can move to any of the four neighboring cells.
   - The distance between two cells \((x_1, y_1)\) and \((x_2, y_2)\) in taxicab geometry is given by:
     \[
     d((x_1, y_1), (x_2, y_2)) = |x_1 - x_2| + |y_1 - y_2|
     \]

3. **Analyze the parity of the starting positions:**
   - The initial position of the policeman is \((1,1)\), which has coordinates of the same parity (both odd).
   - The position \((i,j)\) can have either the same parity (both coordinates are even or both are odd) or different parity (one coordinate is odd and the other is even).

4. **Case 1: \((i,j)\) have the same parity:**
   - If \((i,j)\) have the same parity, then the sum \(i + j\) is even.
   - The initial position \((1,1)\) also has an even sum \(1 + 1 = 2\).
   - When both the policeman and the thief move, they maintain their parity. Thus, the parity of their positions remains the same after each move.
   - The policeman can always move in such a way to reduce the taxicab distance to 1 or 2, eventually cornering the thief.

5. **Case 2: \((i,j)\) have different parity:**
   - If \((i,j)\) have different parity, then the sum \(i + j\) is odd.
   - The initial position \((1,1)\) has an even sum.
   - After the thief's first move, the parity of the thief's position changes, making the sum even.
   - After the policeman's move, the parity of the policeman's position changes, making the sum odd again.
   - Therefore, the parity of the positions alternates, and the taxicab distance between the policeman and the thief remains odd after each of the policeman's moves.
   - Since the distance is always odd, the policeman can never catch the thief.

Conclusion:
The policeman can catch the thief if and only if the coordinates \((i,j)\) of the thief have the same parity.

The final answer is \( \boxed{ (i,j) } \) have the same parity.