Mongolian Mathematical Olympiad

Without loss of generality, we can assume that $1\le m\le n$. We define an "excellent" triangle in the partition as one that has exactly two good sides. An excellent triangle thus has two sides of length 1 and forms a right triangle. When $n\ge 2$, a partition of the $2\times n$ grid contains no excellent triangles, whereas a $1\times n$ grid partition contains 2 excellent triangles. Therefore, if $m,n\ge 2$ and $mn$ is even, there exists a partition without any excellent triangles. Additionally, if $m=1$ or $mn$ is odd, there exists a partition with 2 excellent triangles.

Now we show that if $m=1$ or the product $mn$ is odd, then there exists a partition with at least 2 excellent triangles. We define a side as "bad" if it is not good, and a triangle as "bad" if it is not excellent. A bad triangle has two bad sides, and we refer to the segment connecting the midpoints of these bad sides as the "main" segment. Note that bad sides do not pass through integer vertex coordinates, except within their corresponding triangles.

The main segments of two neighboring bad triangles are connected, forming a chain. These main segments are good and change direction only at the center of unit grids. Therefore, any closed chain formed passes through an even number of unit grids. At the end of any non-closed chain, there must exist two excellent triangles.

If $m=1$, there is no closed chain, and if $mn$ is odd, a closed chain exists. Therefore, there are at least 2 excellent triangles.

Remark. Alternatively, one can consider areas of good triangles.