On a rectangular board with $m$ rows and $n$ columns, where $m\leq n$, some squares are coloured black in a way that no two rows are alike. Find the biggest integer $k$ such that for every possible colouring to start with one can always color $k$ columns entirely red in such a way that no two rows are still alike.
1. **Define the problem and notation:**
   We are given a rectangular board with \( m \) rows and \( n \) columns, where \( m \leq n \). Some squares are colored black (denoted by 1) and others are white (denoted by 0). The goal is to find the largest integer \( k \) such that for any initial coloring, we can always color \( k \) columns entirely red (effectively removing them) in such a way that no two rows are still alike.

2. **Initial observation:**
   Coloring a column red corresponds to removing that column from consideration. We need to ensure that after removing \( k \) columns, the remaining columns still have unique row patterns.

3. **Upper bound for \( k \):**
   To show that anything greater than \( n - m + 1 \) doesn't work, consider the matrix \( X \) where \( x_{ij} = 1 \) if and only if \( i = j \). This is essentially an identity matrix of size \( m \times m \) embedded in a larger \( m \times n \) matrix. If we remove more than \( n - m + 1 \) columns, we are left with fewer than \( m \) columns. In this case, by the pigeonhole principle, at least two rows must be identical, which contradicts the requirement that no two rows are alike.

4. **Lower bound for \( k \):**
   To show that \( n - m + 1 \) indeed works, we need to demonstrate that the difference in the rows can be captured in \( m \) columns. We can use induction on \( m \):

   - **Base case:** For \( m = 1 \), any \( n \geq 1 \) will work since there is only one row, and it is trivially unique.
   - **Inductive step:** Assume that for \( m \) rows, we can always find \( n - m + 1 \) columns to remove such that the remaining rows are unique. Now consider \( m + 1 \) rows. We need to show that we can remove \( n - (m + 1) + 1 = n - m \) columns and still have unique rows.

     Given \( m + 1 \) rows, there are at most \( 2^{m+1} \) possible row patterns. Since no two rows are alike initially, we have exactly \( m + 1 \) unique patterns. By the inductive hypothesis, we can remove \( n - m \) columns from the first \( m \) rows and still have unique rows. The remaining \( m + 1 \)th row must also be unique among the remaining rows, as it was initially unique.

5. **Conclusion:**
   Therefore, the largest integer \( k \) such that for any initial coloring, we can always color \( k \) columns entirely red and still have unique rows is \( n - m + 1 \).

The final answer is \( \boxed{n - m + 1} \).