A $12\times 12$ board is divided into 144 unit squares by drawing lines parallel to the sides. Two rooks placed on two unit squares are said to be non-attacking if they are not in the same column or same row. Find the least number $N$ such that if $N$ rooks are placed on the board, one rook per square, we can always find 7 rooks such that no two are attacking each other.

We have a $12\times 12$ board. Rooks are non-attacking if no two are in the same row or column.
We want the minimum number $N$ such that placing $N$ rooks guarantees a set of 7 pairwise non-attacking rooks.

By Dilworth's Theorem or the bipartite matching version of the Pigeonhole Principle, the maximum number of rooks we can place without having 7 pairwise non-attacking rooks is achieved by completely filling $6$ rows (or columns) with rooks.
If we fill 6 rows, we place $6\times 12=72$ rooks. In this configuration, we can choose at most 6 pairwise non-attacking rooks (since we can choose at most one rook from each of the 6 filled rows).

If we place $73$ rooks, by the Pigeonhole Principle and Hall's Marriage Theorem, we are guaranteed to find at least 7 pairwise non-attacking rooks.
Thus, the minimum number $N$ is $73$.