【Analysis】As shown in the figure, connect $\mathrm{HF}$. Let's assume the side length of the two squares is $\mathrm{a}$. From the given, $\mathrm{GH}=\mathrm{HE}=\frac{1}{2} \mathrm{a}, \mathrm{DF}=\frac{1}{3} \mathrm{a}, \mathrm{FC}=\frac{2}{3} \mathrm{a}$. Since $\mathrm{GM} / / \mathrm{DF}$, we have $\frac{G M}{D F}=\frac{A G}{A D}=\frac{1}{2} \Rightarrow G M=\frac{1}{6} a$.

Therefore, $\mathrm{MH}=\mathrm{GH}-\mathrm{GM}=\frac{1}{3} \mathrm{a}$.
By the hourglass theorem, $\frac{H I}{I D}=\frac{M H}{D F}=1 \Rightarrow H I=I D$.
So $S_{H I F}=\frac{1}{2} S_{H D F}=\frac{1}{2} \times \frac{1}{3} S_{H D C}=\frac{1}{2} \times \frac{1}{3} \times \frac{1}{2} S_{G E C D}=\frac{1}{12} m$.

Since $\mathrm{EN} / / \mathrm{CF}$, we have $\frac{E N}{C F}=\frac{B E}{B C}=\frac{1}{2} \Rightarrow E N=\frac{1}{3} a$.
So $\mathrm{NH}=\mathrm{EH}-\mathrm{EN}=\frac{1}{6} \mathrm{a}$.
By the hourglass theorem, $\frac{H J}{J C}=\frac{H N}{C F}=\frac{1}{4} \Rightarrow H J=\frac{1}{4} \times J C$.
So $S_{H J F}=\frac{1}{5} S_{H C F}=\frac{1}{5} \times \frac{2}{3} S_{H D C}=\frac{1}{5} \times \frac{2}{3} \times \frac{1}{2} S_{G E C D}=\frac{1}{15} \mathrm{~m}$.
Therefore, $n=\frac{1}{12} m+\frac{1}{15} m=\frac{3}{20} m$.
Since $\mathrm{m}, \mathrm{n}$ are both positive integers, $\mathrm{m}$ must be a multiple of 20, i.e., $\mathrm{m}$ contains the prime factors $2$ and $5$. Also, since $\mathrm{m}$ has 9 divisors, $\mathrm{m}=2^{2} \times 5^{2}=100$.
Therefore, the side length of the square $\mathrm{ABEG}$ is 10 cm.