I'll prove the existence of $2n-1$ L. Color columns alterantively, black and white. Each 2X2 square and S covers two white and two black squares. And hebce the difference between the number of black covered squares and white ones is 0. But there are $2n-1$ more white squares. Hence there must be at least $2n-1$ L pieces.

The problem can be solved in a similar way. Coloring the board in chesboard manner we find 4n-1 more white squares. And hence we need $4n-1$ L shaped tiles.

xirti wrote: The problem can be solved in a similar way. Coloring the board in chesboard manner we find 4n-1 more white squares. And hence we need $4n-1$ L shaped tiles. I don't think there will be 4n-1 more white squares.

Yes, actually there will be 1 more white square. So we have to use another colouring. Here I give one solution: Let's number the rows $1,2,...,2n-1$ and the columns $1,...,2n-1$. Let's colour the cells $(i,j)$ where both $i$ and $j$ are odd in black and the others in white. So we will have $n^2$ black cells and $3n^2 -4n +1$ white cells. Now no tile can cover two black tiles. Any $L$-tile covers 1black, 2white squares or 3 white squares. Any other tile covers 1black and 3white squares exactly! We finnaly cover $B=n^2$ black squares and $W=3n^2 -4n +1$ white squares. Consider the sum: $3B-W=4n-1$ Each $L$-tile adds $+1$ or $-3$ to the sum, while the other tiles don't change it. So there must be at least $4n-1$ $L$-tiles.

btw, more interesting question is when such covering exist