Let $m$ and $n$ be positive integers. Some squares of an $m \times n$ board are coloured red. A sequence $a_1, a_2, \ldots , a_{2r}$ of $2r \ge 4$ pairwise distinct red squares is called a [i]bishop circuit[/i] if for every $k \in \{1, \ldots , 2r \}$, the squares $a_k$ and $a_{k+1}$ lie on a diagonal, but the squares $a_k$ and $a_{k+2}$ do not lie on a diagonal (here $a_{2r+1}=a_1$ and $a_{2r+2}=a_2$).
In terms of $m$ and $n$, determine the maximum possible number of red squares on an $m \times n$ board without a bishop circuit.
([i]Remark.[/i] Two squares lie on a diagonal if the line passing through their centres intersects the sides of the board at an angle of $45^\circ$.)
1. **Initial Considerations:**
   - For a \(1 \times n\) or \(n \times 1\) board, the maximum number of red squares without forming a bishop circuit is clearly \(n\). This is because there are no diagonals with more than one square, so no bishop circuit can be formed.
   - For \(m, n > 1\), we need to consider the structure of the board more carefully.

2. **Graph Representation:**
   - Consider a graph \(G\) where each vertex represents a red square on the \(m \times n\) board.
   - Draw an edge between two vertices if the corresponding squares lie on the same diagonal and there are no other red squares between them.

3. **Cycle and Tree Structure:**
   - A bishop circuit corresponds to a cycle in the graph \(G\).
   - To avoid a bishop circuit, \(G\) must be a forest (a collection of trees).
   - A tree with \(N\) vertices has \(N-1\) edges. If \(G\) is a forest with \(i\) connected components, it has \(N-i\) edges.

4. **Coloring Argument:**
   - Color the board like a chessboard with alternating black and white squares.
   - No two black squares are adjacent on a diagonal, and no two white squares are adjacent on a diagonal.
   - This implies that the graph \(G\) has at least two connected components (one for black squares and one for white squares).

5. **Diagonal Counting:**
   - Enumerate the rows of the board as \(1, 2, \ldots, m\) and the columns as \(m+1, m+2, \ldots, m+n\).
   - Define an increasing diagonal as one that starts on the left or bottom side and ends on the top or right side.
   - Define a decreasing diagonal as one that starts on the left or top side and ends on the right or bottom side.

6. **Double Counting:**
   - Let \(a_i\) be the number of red squares on the \(i\)-th increasing diagonal.
   - Let \(b_i\) be the number of red squares on the \(i\)-th decreasing diagonal.
   - By double counting, \(\sum a_i = \sum b_i = N\), where \(N\) is the total number of red squares.
   - The number of edges in the graph for each diagonal is \(a_i - 1\) (or \(b_i - 1\)).

7. **Edge Counting:**
   - The total number of edges in the graph is \(\sum_{i=1}^{m+n-1} (a_i - 1) + \sum_{i=1}^{m+n-1} (b_i - 1)\).
   - This simplifies to \(2N - 2(m+n-1)\).

8. **Inequality:**
   - Since \(G\) is a forest with at least two components, the number of edges is at most \(N - 2\).
   - Therefore, \(2N - 2(m+n-1) \leq N - 2\).
   - Simplifying, we get \(N \leq 2m + 2n - 4\).

9. **Construction:**
   - To achieve \(N = 2m + 2n - 4\), color all squares in the first and last columns and the last two rows.
   - This configuration avoids forming a bishop circuit and uses \(2m + 2n - 4\) red squares.

\[
\boxed{2m + 2n - 4}
\]