We claim that the answer is all $(m,n)$ such that $4|n$ or $4|m$, and $m,n \neq 1$. Assign to the grid the numbers: $a_{jk}=i^{j+k} (0 \le j \le m-1, 0 \le k \le n-1)$, where $i^2=-1$. Trivial calculations show that the sum of numbers in all 6 types of tiles is $0$. Thus, the sum of all numbers on the grid is $0$, i.e. $\sum_{k=0}^{m-1}i^k \cdot \sum_{k=0}^{n-1}i^k = 0$, which implies $4|n$ or $4|m$. Now, clearly a $4 \times 1$ cannot be tiled. And every integer greater than or equal to $2$, can be represented in the form $2x+3y$ for some non-negative integers $x,y$, and the following tilings are enough. (Sorry for the rusty diagram, the $2\times 4$ one should be flipped.