Let $n$ be a positive integer. Find the smallest positive integer $k$ with the property that for any colouring nof the squares of a $2n$ by $k$ chessboard with $n$ colours, there are $2$ columns and $2$ rows such that the $4$ squares in their intersections have the same colour.
1. **Define the problem in terms of graph theory:**
   - Consider a $2n \times k$ chessboard where each cell is colored with one of $n$ colors.
   - We need to find the smallest $k$ such that any coloring of the chessboard guarantees the existence of two rows and two columns where the four intersection cells have the same color.

2. **Graph representation:**
   - Associate each of the $2n$ rows with a vertex of the complete graph $K_{2n}$.
   - Each pair of cells in the same column and with the same color corresponds to assigning that color to one edge of $K_{2n}$.
   - For each edge, each color is allowed to occur once. Therefore, there can be at most $n \cdot \binom{2n}{2}$ such pairs.

3. **Counting pairs:**
   - The number of edges in $K_{2n}$ is $\binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1)$.
   - Since each color can be used for each edge exactly once, the total number of such pairs is $n \cdot n(2n-1) = n^2(2n-1)$.

4. **Column requirement:**
   - Each column must contain at least $n$ such pairs (since there are $2n$ rows and each color can appear at most once in each column).
   - Therefore, if there are more than $k = n(2n-1)$ columns, the property must hold.

5. **Construction for $k = n(2n-1)$:**
   - For $k = n(2n-1)$, we can construct a coloring without the property.
   - Start with a 1-factor of $K_{2n}$ having edges $(1,2), (3,4), \ldots, (2n-1,2n)$. Assign color $i$ to edge $(2i-1,2i)$ for $i = 1, 2, \ldots, n$. This forms the first column.
   - Change the labels $1, \ldots, 2n-1$ cyclically modulo $2n-1$ but leave $2n$ unchanged. This yields $2n-2$ other disjoint (multicolored) 1-factors, each forming another column.
   - Take $n$ instances of the whole matrix of these $2n-1$ columns such that in each one the $n$ colors are cyclically permuted with respect to the first instance.
   - This results in a board of $n(2n-1)$ columns without a monochromatic rectangle.

6. **Conclusion:**
   - The smallest $k$ such that any coloring of the $2n \times k$ chessboard with $n$ colors guarantees the existence of two rows and two columns where the four intersection cells have the same color is $k = n(2n-1) + 1$.

The final answer is $\boxed{n(2n-1) + 1}$.