The $m \times n$ grid is formed by $m+1$ horizontal and $n+1$ vertical lines. Number the horizontal lines with numbers from $1$ to $m+1$ and the vertical lines with numbers from $1$ to $n+1$. Rectangles with odd side lengths arise if and only if two horizontal lines with different parity and two vertical lines with different parity intersect.

Assume that at least one of the numbers $m$ and $n$ is even. We can assume without loss of generality that $m = 2k$. Then there are exactly $k+1$ odd-numbered and $k$ even-numbered horizontal lines and thus there are $k(k+1)$ pairs of lines of different parity. But this means that overall the number of rectangles with odd side lengths is even and cannot be $225$. Therefore $m$ and $n$ are both odd numbers.

Let now $m = 2k-1$ and $n = 2l-1$. Then we have exactly $k$ even-numbered and $k$ odd-numbered horizontal lines and $l$ even-numbered and $l$ odd-numbered vertical lines. Overall it is possible to form $k \cdot k \cdot l \cdot l = (kl)^2$ rectangles with odd side lengths. From $(kl)^2 = 225$ we get $kl = 15$. The solutions are $k=1, l=15$ or $k=3, l=5$ (or vice versa). So $m=1, n=29$ or $m=5, n=9$ (or vice versa).