Given a rectangular grid, split into $m\times n$ squares, a colouring of the squares in two colours (black and white) is called valid if it satisfies the following conditions:
[list]
[*]All squares touching the border of the grid are coloured black.
[*]No four squares forming a $2\times 2$ square are coloured in the same colour.
[*]No four squares forming a $2\times 2$ square are coloured in such a way that only diagonally touching
squares have the same colour.[/list]
Which grid sizes $m\times n$ (with $m,n\ge 3$) have a valid colouring?
To determine which grid sizes \( m \times n \) (with \( m, n \geq 3 \)) have a valid colouring, we need to analyze the given conditions and derive the necessary constraints.

1. **Condition Analysis**:
   - All squares touching the border of the grid are coloured black.
   - No four squares forming a \( 2 \times 2 \) square are coloured in the same colour.
   - No four squares forming a \( 2 \times 2 \) square are coloured in such a way that only diagonally touching squares have the same colour.

2. **Odd Dimensions**:
   - If at least one of \( m \) or \( n \) is odd, we can construct a valid colouring by alternating colours in a checkerboard pattern. This ensures that no \( 2 \times 2 \) square will have the same colour in all four squares or only diagonally touching squares with the same colour.
   - For example, if \( m \) is odd and \( n \) is even, we can start with black on the border and alternate colours. This pattern will satisfy all conditions.

3. **Even Dimensions**:
   - If both \( m \) and \( n \) are even, we need to show that a valid colouring is impossible.
   - Consider the grid as a graph where each square is a vertex, and an edge exists between vertices if they are adjacent and have different colours.
   - The conditions imply that every interior vertex (square) must have exactly two edges (segments) joined to it, forming a bipartite 2-regular graph.
   - A bipartite 2-regular graph with an odd number of vertices is impossible because it would require an even number of vertices to maintain the bipartite property and the 2-regularity.

4. **Graph Theory Argument**:
   - For an \( (m-1) \times (n-1) \) grid of interior vertices, if both \( m \) and \( n \) are even, the number of interior vertices is \((m-1)(n-1)\), which is odd.
   - A bipartite 2-regular graph with an odd number of vertices cannot exist because it would require an even number of vertices to maintain the bipartite property and the 2-regularity.

Therefore, the grid sizes \( m \times n \) that have a valid colouring are those where at least one of \( m \) or \( n \) is odd.

The final answer is \( \boxed{\text{At least one of } m \text{ or } n \text{ must be odd}} \)