4. As shown in Figure 2, two congruent equilateral triangles $\triangle A_{1} B_{1} C_{1}$ and $\triangle A_{2} B_{2} C_{2}$ with side lengths as positive integers have their centers coinciding, and satisfy $A_{1} B_{1} \perp A_{2} C_{2}$. If the area of the hexagon $A B C D E F$ is $S=\frac{1}{m}-\frac{\sqrt{3}}{n}$, where $m$ and $n$ are rational numbers. Then the value of $\frac{m}{n}$ is $\qquad$
4. $\frac{1}{3}$.

By symmetry, we know that $A_{1} B_{1} \perp A_{2} C_{2}, B_{1} C_{1} \perp A_{2} B_{2}, C_{1} A_{1} \perp B_{2} C_{2}$, which means the right triangles $\triangle A_{1} A F, \operatorname{Rt} \triangle A_{2} A B, \operatorname{Rt} \triangle B_{1} C B, \operatorname{Rt} \triangle B_{2} C D, \operatorname{Rt} \triangle C_{1} E D, \operatorname{Rt} \triangle C_{2} E F$ are all congruent. Therefore, we examine $\triangle A_{1} A F$.

Let $A_{1} B_{1}=a\left(a \in \mathbf{N}_{+}\right), A A_{1}=x$. Then $A F=\sqrt{3} x$, $A_{1} F=2 x$. We have $x+\sqrt{3} x+2 x=a$. Solving for $x$ gives $x=\frac{a}{3+\sqrt{3}}$.
Thus, $S_{\triangle A_{1} A F}=\frac{\sqrt{3}}{2} x^{2}=\frac{\sqrt{3}}{2}\left(\frac{a}{3+\sqrt{3}}\right)^{2}$.
Then $S=\frac{\sqrt{3}}{4} a^{2}-3 S_{\triangle A_{1} A F}=\frac{3}{4} a^{2}-\frac{\sqrt{3}}{4} a^{2}$.
Given that $S=\frac{1}{m}-\frac{\sqrt{3}}{n}$ and $a$ is a positive integer, $m, n$ are rational numbers, we get $m=\frac{4}{3 a^{2}}, n=\frac{4}{a^{2}}$.