Problem 1387

1. Initial Setup: Consider a $m \times (m-1)$ board, which has $m+1$ horizontal lines and $m$ vertical lines. A stone is placed at an intersection on the lowest horizontal line.

2. First Move: The first player moves the stone up to point $A$. This move effectively removes the bottom horizontal line from consideration, transforming the board into an $m \times m$ square.

3. **Rectangle $\Box ABCD$**: Consider a rectangle $\Box ABCD$ inscribed in the $m \times m$ square, where the slopes of the lines forming the rectangle are either $1$ or $-1$. This rectangle is formed by the diagonals of the square.

4. Movement Strategy: The first player can always keep the stone inside (including the borders of) $\Box ABCD$. This is because the first player can always move the stone to a neighboring intersection along a segment that has not been used before.

5. Vertex Degrees: In the rectangle $\Box ABCD$, every point except the vertices $B$, $C$, and $D$ has an even degree. This means that each of these points can be entered and exited an even number of times, ensuring that the stone can continue to move.

6. Parity Argument: The first player can always move to vertices $B$, $C$, or $D$ first due to the parity of moves. Since the first player starts the game, they will always have the advantage of making the first move to these critical points.

7. Winning Condition: By moving to $B$, $C$, or $D$ first, the first player ensures that the second player will eventually be unable to make a move, thus securing the first player's victory.

$$\boxed{\text{There is a winning strategy for the first player.}}$$