A piece is placed in the lower left-corner cell of the $15 \times 15$ board. It can move to the cells that are adjacent to the sides or the corners of its current cell. It must also alternate between horizontal and diagonal moves $($the first move must be diagonal$).$ What is the maximum number of moves it can make without stepping on the same cell twice$?$
1. **Generalization and Initial Claim:**
   - We start by generalizing the problem to an \( n \times n \) board, where \( n \) is odd.
   - We claim that the maximum number of moves is \( (n-1)^2 \).

2. **Example for \( 5 \times 5 \) Board:**
   - Consider a \( 5 \times 5 \) board. We can construct a sequence of moves as follows:
     \[
     \begin{bmatrix}
     & 8 & 9 & & \\
     7 & 6 & 11 & 10 & \\
     5 & 4 & 13 & 12 & \\
     3 & 2 & 15 & 14 & \\
     1 & & & 16 & 17 \\
     \end{bmatrix}
     \]
   - This sequence shows 16 moves, which is \( (5-1)^2 = 16 \).

3. **Generalization of the Example:**
   - The pattern observed in the \( 5 \times 5 \) board can be generalized to any \( n \times n \) board.
   - The number of moves is given by:
     \[
     (n-1) + (n-1) + \ldots + 1 - 1 = (n-1)^2
     \]

4. **Bounding the Number of Moves:**
   - We use a chessboard coloring of the board, with the corner squares being black.
   - Define a *brick* as a shape consisting of two squares that share one corner.
   - Bricks can be of two types:
     - *Type A*: Consisting of only black squares.
     - *Type B*: Consisting of only white squares.

5. **Path Analysis:**
   - The path alternates between Type A and Type B bricks, starting with a Type A brick since the lower-left square is black.
   - Let \( X \) be the number of Type A bricks and \( Y \) be the number of Type B bricks.
   - We have \( Y \in \{X, X-1\} \).

6. **Calculating the Number of Moves:**
   - In the first case (\( Y = X \)), the number of moves is:
     \[
     2(X + Y) + 1 - 1 = 4X
     \]
   - In the second case (\( Y = X-1 \)), the number of moves is:
     \[
     2(X + Y) + 1 - 1 = 4X - 2
     \]
   - Thus, the total number of moves is always less than or equal to \( 4X \).

7. **Claim:**
   - The total number of Type A bricks in the path is less than or equal to \( \frac{(n-1)^2}{4} \).

8. **Proof of the Claim:**
   - Consider only the black squares of the board.
   - There are \( \frac{(n-1)^2}{4} \) internal black squares (excluding the sides of the board).
   - Each Type A brick must have exactly one square belonging to these internal black squares.
   - Since the path has no loops, at most one Type A brick can be "rooted" at each internal black square.
   - Therefore, the number of Type A bricks is at most \( \frac{(n-1)^2}{4} \).

9. **Conclusion:**
   - The maximum number of moves is \( (n-1)^2 \).

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