For $n$ an odd positive integer, the unit squares of an $n\times n$ chessboard are coloured alternately black and white, with the four corners coloured black. A it tromino is an $L$-shape formed by three connected unit squares. For which values of $n$ is it possible to cover all the black squares with non-overlapping trominos?  When it is possible, what is the minimum number of trominos needed?
1. **Express \( n \) in terms of \( k \):**
   Let \( n = 2k + 1 \), where \( k \) is a non-negative integer. This ensures \( n \) is an odd positive integer.

2. **Initial claim:**
   We claim that for all odd \( n \geq 7 \), it is possible to cover all the black squares with non-overlapping trominos, and the minimum number of trominos required is \( (k + 1)^2 \).

3. **Constructing the board:**
   Consider the \( n \times n \) chessboard. We can divide this board into four rectangles:
   - Two \( k \times (k + 1) \) rectangles.
   - Two \( (k + 1) \times k \) rectangles.
   
   These rectangles leave the center square uncovered.

4. **Counting trominos for odd \( k \):**
   If \( k \) is odd, the number of trominos required to cover one \( k \times (k + 1) \) rectangle is \( \left(\frac{k + 1}{2}\right)^2 \). Since there are four such rectangles, the total number of trominos is:
   \[
   4 \times \left(\frac{k + 1}{2}\right)^2 = (k + 1)^2
   \]

5. **Counting trominos for even \( k \):**
   If \( k \) is even, the number of trominos required to cover one \( k \times (k + 1) \) rectangle is \( \left(\frac{k}{2}\right)^2 \). Again, since there are four such rectangles, the total number of trominos is:
   \[
   4 \times \left(\frac{k}{2}\right)^2 = (k + 1)^2
   \]

6. **Induction hypothesis:**
   Assume for a \( (2k - 1) \times (2k - 1) \) board, it takes \( k^2 \) trominos to cover all the black squares.

7. **Induction step:**
   For a \( (2k + 1) \times (2k + 1) \) board, we have an extra \( 8k \) squares, which can be divided into:
   - A \( 2 \times (2k + 1) \) rectangle.
   - A \( (2k - 1) \times 2 \) rectangle.
   
   The minimum number of trominos required to tile a \( 2 \times (2k + 1) \) rectangle is \( k + 1 \), and for a \( (2k - 1) \times 2 \) rectangle is \( k \). Thus, the new minimum number of trominos is:
   \[
   k^2 + (k + 1) + k = (k + 1)^2
   \]

8. **Conclusion:**
   By induction, we have shown that for all odd \( n \geq 7 \), it is possible to cover all the black squares with non-overlapping trominos, and the minimum number of trominos required is \( (k + 1)^2 \).

The final answer is \( \boxed{ (k + 1)^2 } \)