Problem 1276

1. Coloring the Chessboard:
We color the chessboard in a repeating pattern of 1, 2, 3, and 4 as follows:
$$\begin{array}{cccc} 1 & 2 & 1 & 2 & \ldots \\ 4 & 3 & 4 & 3 & \ldots \\ 1 & 2 & 1 & 2 & \ldots \\ 4 & 3 & 4 & 3 & \ldots \\ \vdots & \vdots & \vdots & \vdots & \ddots \\ \end{array}$$

2. Analyzing Moves:
- (b) Move: Moves the stone to a square sharing only one common vertex with the current square. This means the stone moves to a square of the same parity but different color.
- (a) Move: Moves the stone to a square sharing a common side with the current square. This means the stone moves to a square of different parity.

3. Path Analysis:
Consider a sequence of moves in groups of 4 squares: A, B, C, D.
- A to B is a (b) move: A and B have the same parity but different colors.
- B to C is an (a) move: B and C have different parities.
- C to D is a (b) move: C and D have the same parity but different colors.

This implies:
- $A$ and $B$ have the same parity.
- $C$ and $D$ have the same parity.
- $B$ and $C$ have different parities.
- $A \neq B$ and $C \neq D$.

Therefore, $\{A, B, C, D\} = \{1, 2, 3, 4\}$. Every 4 consecutive squares must include one of each color.

4. **Odd $n$ Analysis:**
Let $n = 2k + 1$. The total number of squares is $n^2 = (2k+1)^2 = 4k^2 + 4k + 1$.
- We can decompose this path into $(k^2 + k)$ blocks of 4 squares and one final square.
- Each block of 4 squares must contain one of each color (1, 2, 3, 4).

However, there are only $k^2$ squares of color 4 on the board. Since we need at least $k^2 + k$ squares of each color, it is impossible to visit every square exactly once for odd $n \geq 3$.

5. **Even $n$ Analysis:**
For even $n$, let $n = 2k$. The total number of squares is $n^2 = (2k)^2 = 4k^2$.
- We can decompose this path into $k^2$ blocks of 4 squares.
- Each block of 4 squares must contain one of each color (1, 2, 3, 4).

Since there are exactly $k^2$ squares of each color, it is possible to visit every square exactly once for even $n$.

Conclusion:
For the stone to visit every square exactly once, $n$ must be even.

The final answer is $\boxed{ n }$ must be even.