Solution:

Let $n=2024$ and $k=1000$. We claim that the maximum number of squares that can be coloured in this way is $k(2n-k)$, which evaluates to $3048000$.

Indeed, call a row/column bad if it has at least one red square. After the first move, there are exactly $k+1$ bad rows and columns: if a row was picked, then that row and the $k$ columns corresponding to the chosen squares are all bad. Any subsequent move increases the number of bad rows/columns by at least $1$. Since there are only $2n$ rows and columns, we can make at most $2n-(k+1)$ moves after the first one, and so at most $2n-k$ moves can be made in total. Thus we can have at most $k(2n-k)$ red squares.

To prove this is achievable, let's choose each of the $n$ columns in the first $n$ moves, and colour the top $k$ cells in these columns. Then, the bottom $n-k$ rows are still uncoloured, so we can make $n-k$ more moves, colouring $k(n+n-k)$ cells in total.