Here is a solution different from the writer’s.

Let us divide \(3N\) black squares into \(N\) components, most of which are of size \(1\). After making one big connected component of size approximately \(2N\) and putting it diagonally, we can reach the following solution by some tweaks, mainly involving the size-\(1\) components.

N = 6    N = 7     N = 8
#..##.   #...##.   #....##.
##...#   ##....#   ##.....#
.##..#   .##...#   .##....#
#.##..   #.##...   #.##....
.#.##.   .#.##..   .#.##...
..#.##   ..#.##.   ..#.##..
         ...#.##   ...#.##.
                   ....#.##