$N = (k - 1)^2 n$.

Choose $(k-1)^2$ rows of the grid: colour the first $k-1$ rows in colour $c_1$; the second $k-1$ rows in colour $c_2$; ...; the last $k-1$ rows in colour $c_{k-1}$. Altogether, $(k-1)^2 n$ squares are coloured. For any coloured $k$-set, as there are only $k-1$ colours, some two squares must have the same colour. On the other hand, if all the squares in this $k$-set have the same colour, then by definition they are in $k$ different rows, yet there are only $k-1$ rows in that colour, a contradiction. This implies that some two squares have different colours. Therefore, $N \ge (k-1)^2 n$.

In a $k$-set, if all the squares have the same colour, call it a mono $k$-set; if the squares have distinct colours, call it a poly $k$-set. We assert that for any colouring of $(k-1)^2n + 1$ squares, there must exist a mono $k$-set or a poly $k$-set. This will give $N \le (k-1)^2n$ and the conclusion. First, we need a lemma.

Lemma In an $n \times n$ grid, among any $(m-1)n + 1$ squares, $1 \le m \le n$, there exists an $m$-set.

Proof of lemma Divide the squares of the $n \times n$ grid into $n$ groups, such that the square in the $i$th row and $j$th column is in group $a$ ($1 \le i, j, a \le n$) if and only if $i - j \equiv a \pmod{n}$. Notice that for each group, the squares are in distinct rows and distinct columns. By the pigeonhole principle, among any $(m-1)n + 1$ squares, $m$ of them are in a group, and they form an $m$-set.

For the original problem, assume that $(k-1)^2n+1$ squares are coloured in a certain way. According to the lemma, there exists a $((k-1)^2+1)$-set, call it $A$. If there are $k-1$ or fewer colours in $A$, then by the pigeonhole principle some $k$ squares are in the same colour, and they form a mono $k$-set; if there are $k$ or more colours in $A$, then choose $k$ squares of distinct colours, and they form a poly $k$-set.

Therefore, the largest $N$ is $(k-1)^2n$.

Alternative proof of lemma Choose any $(m-1)n+1$ squares and colour them black. Take $m$ rows with the most black squares: say they are row $1, 2, \dots, m$, with $x_1, x_2, \dots, x_m$ black squares, respectively, and the other rows have $x_{m+1}, \dots, x_n$ black squares, respectively. If there exists $l$, $1 \le l \le m$, such that in the first $m$ rows, black squares of some $l$ rows are distributed in $l-1$ (or fewer) columns, say, black squares of row $1, \dots, l$ are all in the first $l-1$ columns. Then, $x_1, \dots, x_l \le l-1$; for $i > m$, $x_i \le l-1$ as well (since there are fewer black squares in those rows). Hence,
$$\begin{align*} (m-1)n + 1 &= \sum_{i=1}^{n} x_i \le l(l-1) + (m-l)n + (n-m)(l-1) \\ &= mn - n + m - l(m+1-l) \\ &\le mn - n + m - m = (m-1)n, \end{align*}$$
which is a contradiction. It follows that for any $l$ of the first $m$ rows, the black squares are located in at least $l$ columns. By Hall's marriage theorem, there are $m$ black squares in distinct rows and distinct columns, which is an $m$-set. The lemma is proved. $\square$