Let's look at the board by coloring its columns in an alternating fashion with 3 colors and let $A_i$ be the set of cells with the color $i$: \[ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline 1 & 2 & 3 & 1 & 2 & 3 & 1 & 2 & 3 \\\hline \end{array} \]From this coloring, we get that as we put either of the two tiles the amount of cells covered from each color change their parity or stays the same. Then, the amount of cells from each color on the board has to be of the same parity. Suppose $3\nmid m$, then, $A_1$ has n mores cells than $A_3$, but $A_1\equiv A_3 (mod\ 2)\Longrightarrow 2\mid n$ Analogously, by coloring its rows $3\nmid n\Longrightarrow 2\mid n$, then $3\mid m,n; 2\mid m,n, 6\mid m$ or $6\mid n$ the examples are not that hard, so they are left for the reader