4. As shown in Figure 3, in the square paper piece $A B C D$, $E$ is the midpoint of $B C$. Fold the square so that point $A$ coincides with $E$, and flatten it with the crease being $M N$. Then the ratio of the area of trapezoid $A D M N$ to the area of trapezoid $B C M N$ is $\qquad$
4. $\frac{3}{5}$.

As shown in Figure 5, let the side length of the square be 2. Then $A E=\sqrt{5}$.

Draw a perpendicular from point $M$ to $A B$, intersecting at point $P$.

It is easy to see that $M N$ is the perpendicular bisector of $A E$.
Thus, $\triangle M P N \cong \triangle$
$A B E$
$$
\Rightarrow P N=B E=1 \text {. }
$$

By $\triangle A N T \backsim \triangle A E B$, we get $A N=\frac{5}{4}$. Then $D M=A P=\frac{1}{4}$.
$$
\text { Therefore, } \frac{S_{\text {trapezoid } A D M N}}{S_{\text {trapezoid } B C M N}}=\frac{A N+D M}{M C+N B}=\frac{3}{5} \text {. }
$$