In a $12\times 12$ grid, colour each unit square with either black or white, such that there is at least one black unit square in any $3\times 4$ and $4\times 3$ rectangle bounded by the grid lines. Determine, with proof, the minimum number of black unit squares.
1. **Divide the Grid into Rectangles**:
   - Consider the $12 \times 12$ grid. We can divide this grid into non-overlapping $3 \times 4$ rectangles. Since the grid is $12 \times 12$, we can fit $12$ such $3 \times 4$ rectangles (3 rows and 4 columns of $3 \times 4$ rectangles).

2. **Minimum Number of Black Squares in Each Rectangle**:
   - Each $3 \times 4$ rectangle must contain at least one black square. Therefore, for the $12$ non-overlapping $3 \times 4$ rectangles, we need at least $12$ black squares.

3. **Verification for $4 \times 3$ Rectangles**:
   - Similarly, we can divide the $12 \times 12$ grid into $4 \times 3$ rectangles. Since the grid is $12 \times 12$, we can fit $12$ such $4 \times 3$ rectangles (4 rows and 3 columns of $4 \times 3$ rectangles).
   - Each $4 \times 3$ rectangle must also contain at least one black square. Therefore, for the $12$ non-overlapping $4 \times 3$ rectangles, we need at least $12$ black squares.

4. **Constructing the Grid**:
   - We need to ensure that each $3 \times 4$ and $4 \times 3$ rectangle contains at least one black square. One possible way to achieve this is to place black squares in a pattern that covers both types of rectangles.
   - Consider the following pattern:
     \[
     \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|}
     \hline
     \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space \\
     \hline
     \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space \\
     \hline
     \space & \space & X & \space & \space & \space & X & \space & \space & X & \space & \space \\
     \hline
     \space & \space & \space & \space & \space & X & \space & \space & \space & \space & \space & \space \\
     \hline
     \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space \\
     \hline
     \space & \space & X & \space & \space & \space & \space & \space & \space & X & \space & \space \\
     \hline
     \space & \space & \space & \space & \space & X & \space & \space & X & \space & \space & \space \\
     \hline
     \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space \\
     \hline
     \space & \space & \space & X & \space & \space & \space & \space & \space & \space & \space & \space \\
     \hline
     \space & \space & X & \space & \space & \space & X & \space & \space & X & \space & \space \\
     \hline
     \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space \\
     \hline
     \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space & \space \\
     \hline
     \end{array}
     \]
   - This pattern ensures that each $3 \times 4$ and $4 \times 3$ rectangle contains at least one black square.

5. **Conclusion**:
   - The minimum number of black squares required to ensure that every $3 \times 4$ and $4 \times 3$ rectangle contains at least one black square is $12$.

The final answer is $\boxed{12}$