Poly and Omino play a rectangle tiling game

We assume that $N$ divides the product $RC$, because otherwise it is impossible to tile the $R\times C$ board by $N$-ominoes, so then Omino always wins.

Observe that if Poly wins with $N$-polyominoes at $R'\times C'$ board and $R\ge R'$, $C\ge C'$ then Poly also wins with $N$-polyominoes at $R'\times C'$ board. This is because then the $R\times C$ board with $R'\times C'$ board removed from the corner has a Hamiltonian path of adjacent cells, which can be tiled by $N$-ominoes.

For simplicity we assume that $R\le C$.

Omino wins if:

$N\ge 7$, because then Omino can chose an $N$-mino with $1\times 1$ hole and then it will be impossible to tile a rectangular board using this $N$-mino.

$N>C$, because then there is no tiling of the board using $1\times N$ rectangle.

$N>2R$, because then there is no tiling of the board using some $L$-shaped $N$-omino.

$N=4$ and $R\le 2$, because then Omino can chose this tetromino:

  x
 xxx
 

$N=5$, $R\le 3$, and $C=5$, because then Omino can chose this pentomino:

 x
 xx
  xx
 

$N=6$ and $R\le 3$, because then Omino can chose this hexomino:

  x
  x
 xxxx
 

Poly wins if:

$N\le 3$, because then Poly can place the $N$-omino chosen by Omino at a corner of the board and then tile the remaining space.

$N=4$ and $R\ge 3$. This follows from the observation at the beginning of the answer and the fact that Poly wins at the $3\times 4$ board.

$N=5$, $R\ge 3$, and $RC\ge 20$. This follows from the observation at the beginning of the answer and the facts that Poly wins at the $3\times 10$ and $4\times 5$ boards.

$N=6$, $R\ge 4$, and $C\ge 6$. This follows from the observation at the beginning of the answer and the fact that Poly wins at the $4\times 6$ board.