You can tile with 1, 4, and 6 squares already.
You cannot tile with 2, 3, or 5 squares. For 2 and 3, you can see this because no tile can cover more than one of the corners of the big square. For 5 tiles, there must be one tile in each of the corners of the big square and there must remain at least two edges with uncovered gaps which cannot be filled by a single square.
For any even number $N=2m>2$, you can tile using $2m-1$ squares of size $1\times 1$ along two sides of an $(m-1)\times(m-1)$ square. This gives solutions for $N=4, 6, 8, 10,\ldots$.
Finally, if you have a tiling with $n$ squares, you can make one with $n+3$ squares by cutting one of the tiles into 4 smaller ones. Just apply this once to the above solution and you have $N=7, 9, 11, \ldots$ solved.
Thus, there are tilings for all $N$ except 2, 3, and 5.