A natural number $k$ is said $n$-squared if by colouring the squares of a $2n \times k$ chessboard, in any manner, with $n$ different colours, we can find $4$ separate unit squares of the same colour, the centers of which are vertices of a rectangle having sides parallel to the sides of the board. Determine, in function of $n$, the smallest natural $k$ that is $n$-squared.
1. **Define the problem and notation:**
   We need to determine the smallest natural number \( k \) such that for any coloring of a \( 2n \times k \) chessboard with \( n \) different colors, there exist 4 unit squares of the same color forming a rectangle with sides parallel to the sides of the board.

2. **Prove that \( k = 2n^2 - n + 1 \) is \( n \)-squared:**
   - Consider any column of the \( 2n \times k \) chessboard.
   - In each column, the maximum number of squares that can be colored with different colors is \( n \) (since there are only \( n \) colors).
   - Therefore, there will be at least \( 2n - n = n \) squares in each column that share a color with at least one other square in the same column.
   - This implies that there are at least \( kn \) pairs of squares in the entire board that are in the same column and have the same color.

3. **Averaging argument:**
   - By the pigeonhole principle, since there are \( n \) colors, there must be at least one color that appears in at least \( \frac{kn}{n} = k \) pairs of squares in the same column.

4. **Counting pairs:**
   - The number of possible pairs of squares in the same column is given by \( \binom{2n}{2} = \frac{2n(2n-1)}{2} = n(2n-1) \).
   - For \( k \geq 2n^2 - n + 1 \), the number of pairs \( kn \) exceeds the number of possible pairs \( n(2n-1) \), ensuring that there will be a repetition of the position of a colored pair in at least two columns.

5. **Forming a rectangle:**
   - When there is a repetition of the position of a colored pair in at least two columns, the centers of these four squares will form a rectangle with sides parallel to the sides of the board.

6. **Constructing the coloring:**
   - To show that \( k = 2n^2 - n + 1 \) is the smallest such \( k \), we need to demonstrate that it is possible to color a \( 2n \times (2n-1) \) array for each of the \( n \) colors such that no pair of squares on the same column and colored with the same color repeats on another column.
   - This can be achieved using the method of scheduling a round-robin tournament between \( 2n \) teams in \( 2n-1 \) rounds, ensuring that no two teams play each other more than once.

\(\blacksquare\)

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