4. As shown in Figure 3, given that the area of the equilateral $\triangle A B C$ is $S$, $M$ and $N$ are the midpoints of sides $A B$ and $A C$ respectively, $M E \perp B C$ at $E$, and $E F \perp A C$ at $F$. Let the areas of $\triangle A M N$, $\triangle B E M$, $\triangle E F C$, and quadrilateral $M E F N$ be $S_{1}$, $S_{2}$, $S_{3}$, and $S_{4}$ respectively. Then $\frac{S_{4}}{S_{1}+S_{2}+S_{3}}=$ $\qquad$ .
4. $\frac{11}{21}$.

As shown in Figure 6, connect $B N$, and draw the altitude $A H$. Since $\triangle A B C$ is an equilateral triangle, we know that $B N \perp A C$.

From $\triangle M B E \backsim \triangle A B H$, we get
$$
\frac{S_{2}}{S}=\frac{1}{8},
$$

which means $S_{2}=\frac{1}{8} S$.
Since $\frac{B E}{B H}=\frac{1}{2}$, we have $\frac{C E}{B C}=\frac{3}{4}$.
Also, since $\triangle E C F \backsim \triangle B C N$, we get $\frac{S_{3}}{S}=\frac{9}{32}$, which means $S_{3}=\frac{9}{32} S$.
From $\triangle A M N \backsim \triangle A B C$, we have $\frac{S_{1}}{S}=\frac{1}{4}$, which means $S_{1}=\frac{1}{4} S$.
Therefore, $S_{4}=S-\left(S_{1}+S_{2}+S_{3}\right)=\frac{11}{32} S$. Hence, $\frac{S_{4}}{S_{1}+S_{2}+S_{3}}=\frac{11}{21}$.