Determine all positive integers $ n$ such that it is possible to tile a $ 15 \times n$ board with pieces shaped like this:

[asy]size(100); draw((0,0)--(3,0)); draw((0,1)--(3,1)); draw((0,2)--(1,2)); draw((2,2)--(3,2)); draw((0,0)--(0,2)); draw((1,0)--(1,2)); draw((2,0)--(2,2)); draw((3,0)--(3,2)); draw((5,0)--(6,0)); draw((4,1)--(7,1)); draw((4,2)--(7,2)); draw((5,3)--(6,3)); draw((4,1)--(4,2)); draw((5,0)--(5,3)); draw((6,0)--(6,3)); draw((7,1)--(7,2));[/asy]
To determine all positive integers \( n \) such that it is possible to tile a \( 15 \times n \) board with the given pieces, we need to analyze the shape and tiling possibilities of the pieces.

1. **Understanding the Piece Shape**:
   The given piece is an L-shaped tile that covers exactly 4 unit squares. This means that any tiling of the \( 15 \times n \) board must cover \( 15n \) unit squares, and since each tile covers 4 squares, \( 15n \) must be divisible by 4. Therefore, \( n \) must be such that \( 15n \equiv 0 \pmod{4} \).

2. **Divisibility Condition**:
   Since \( 15 \equiv 3 \pmod{4} \), we need \( 3n \equiv 0 \pmod{4} \). This implies that \( n \) must be a multiple of 4. Therefore, \( n \) must be of the form \( n = 4k \) for some integer \( k \).

3. **Checking Small Values**:
   We need to check small values of \( n \) to see if they can be tiled:
   - \( n = 1 \): \( 15 \times 1 \) board has 15 squares, not divisible by 4.
   - \( n = 2 \): \( 15 \times 2 \) board has 30 squares, not divisible by 4.
   - \( n = 3 \): \( 15 \times 3 \) board has 45 squares, not divisible by 4.
   - \( n = 4 \): \( 15 \times 4 \) board has 60 squares, which is divisible by 4.

4. **General Form**:
   We need to show that for \( n = 4k \), it is possible to tile the \( 15 \times n \) board. We can use the fact that \( n = 3a + 5b \) for non-negative integers \( a \) and \( b \). This is a form of the Frobenius coin problem, where we need to express \( n \) as a linear combination of 3 and 5.

5. **Base Cases and Induction**:
   - For \( n = 3 \), \( n = 5 \), and \( n = 6 \), we can manually verify that these can be tiled.
   - For \( n = 8 \), \( n = 9 \), and \( n = 10 \), we can also verify manually.
   - For \( n \geq 11 \), we can use induction. If \( n \) is achievable, then \( n+1 \) is achievable by adjusting the coefficients \( a \) and \( b \).

6. **Inductive Step**:
   - If \( n = 3a + 5b \) is achievable, then \( n+1 = 3(a+2) + 5(b-1) \) if \( b \geq 1 \).
   - If \( b = 0 \), then \( n = 3a \) and we need to show that \( n+1 = 3(a-3) + 5(2) \) is achievable for \( a > 3 \).

7. **Conclusion**:
   By verifying small cases and using the inductive step, we can conclude that \( n = 1, 2, 4, 7 \) are not achievable, but every other \( n \) is achievable.

The final answer is \( \boxed{ n \neq 1, 2, 4, 7 } \) and all other positive integers are achievable.