Let $n=2k-1$. The black squares consist of $k^2$ "odd squares" (both coordinates odd) and $(k-1)^2$ even squares (both coordinates even). No tromino can cover two odd squares, so we need at least $k^2$ trominoes to cover all black squares. Impossible for $n=3$ and also not for $n=5$, as $9$ trominoes would occupy $27$ squares and $27>25$, so there would be some overlap. For $n=7$, it is not hard to find a covering. (14 of the 16 trominoes will come in pairs making up $2\times 3$ rectangles.) Now by induction, we can construct optimal coverings for $n=9,11,...$: Adding to an $(n-1)^2$ board a hook of width 2 containing $n$ odd and $n-1$ even black squares, and those can be covered by $n$ trominoes as is easily seen.