First we notice that n = 2k because the tetrominoes have a even ammount of total squares. Also we can easily configure them to form a 4x4 square (too lazy to draw), with this we know that the squares of n=4k are possible, now we need to disscard n=4k+2. Assume n=4k+2. We can color the tetrominoes like a chess board in white and black in 2 different ways, one where there is one white square and 3 black, and another the other way around so there is one black and 3 white. It's easy to see that half of the squares in the n x n board are white and the other half is black, therefore there must be an equal ammount of tetrominoes that are colored in one way than in the other, this implies there is a even ammount, so the board must have a total of squares that is a multiple of 8, now (4k+2)(4k+2)=16k^2 + 16k + 4 which is congruent to 4 (mod 8). Therefore only n=4k work.