The answer is $2n^2 - n + 1$.

Consider an $n$-colouring of the $2n \times k$ chessboard. A vertical-pair is a pair of squares in the same column that are coloured the same. In every column there are at least $n$ vertical-pairs. Let $P$ be the total number of vertical-pairs and $P_i$ be the number of vertical-pairs with colour $i$. Then $P = P_1 + \dots + P_n \ge nk$. Thus there is colour $i$ with $P_i \ge k$. There are $\binom{2n}{2} = 2n^2 - n$ pairs of rows. Thus if $k \ge 2n^2 - n + 1$, there is a pair of rows that contains two vertical-pairs with colour $i$.

Next for $k = 2n^2 - n$, exhibit an $n$-colouring where no such sets of 4 squares exists. Note that it suffices to find such an $n$-colouring for the $2n \times (2n-1)$ board. We can then rotate the colours to obtain $n$ of these boards which can then be put together to obtain the requiring $n$-colouring of the $2n \times (2n^2 - n)$ board. For each $i = 1, 2, \dots, 2n-1$, let $A_i = \{(i, 2n-1+i), (i+1, 2n-2+i), \dots, (n-1+i, n+i)\}$, where $2n+k, k > 0$, is taken to be $k$. Note that the pairs in each $A_i$ give a partition of $\{1, 2, \dots, 2n\}$. Moreover, each pair of elements appears in exactly one $A_i$. Now colour the squares of column $i$ using $n$ colours so that the two squares in each pair of $A_i$ receive the same colour and the colours the $2n$ pairs are mutually distinct. This gives an $n$-colouring of the $2n \times 2n - 1$ board with the required property and we are done.