4. As shown in Figure $1, \overrightarrow{A A_{2}}+\overrightarrow{B B_{2}}+\overrightarrow{C C_{2}}=\mathbf{0}$, and $\frac{\left|A B_{1}\right|}{\left|B_{1} C\right|}$ $=\frac{1}{4}$. Then $\frac{\left|A B_{2}\right|}{\left|B_{2} C_{2}\right|}=$ $\qquad$ .

Figure 1
4. $\frac{4}{3}$.
$$
\begin{array}{l}
\text { Given } \overrightarrow{A A_{2}}+\overrightarrow{B B_{2}}+\overrightarrow{C C_{2}}=\mathbf{0}, \\
\overrightarrow{A_{2} B_{2}}+\overrightarrow{B_{2} C_{2}}+\overrightarrow{C_{2} A_{2}}=\mathbf{0} \\
\Rightarrow \overrightarrow{A B_{2}}+\overrightarrow{B C_{2}}+\overrightarrow{C A_{2}}=\mathbf{0} . \\
\text { Let } \overrightarrow{A B_{2}}=k \overrightarrow{B_{2} C_{2}}, \overrightarrow{B C_{2}}=t \overrightarrow{C_{2} A_{2}}, \\
\overrightarrow{C A_{2}}=s \overrightarrow{A_{2} B_{2}} .
\end{array}
$$

Then $k \overrightarrow{B_{2} C_{2}}+t \overrightarrow{C_{2} A_{2}}+s \overrightarrow{A_{2} B_{2}}=\mathbf{0}$, compared with
$$
\overrightarrow{B_{2} C_{2}}+\overrightarrow{C_{2} A_{2}}+\overrightarrow{A_{2} B_{2}}=\mathbf{0}
$$

By the basic theorem of planar vectors, we get
$$
k=s=t .
$$

Draw a line $B_{2} E \parallel B B_{1}$, intersecting $A C$ at point $E$.
$$
\begin{array}{l}
\text { Then } \frac{C E}{E B_{1}}=\frac{C B_{2}}{B_{2} A_{2}}=\frac{A C_{2}}{C_{2} B_{2}}=\frac{A B_{1}}{B_{1} E} \\
\Rightarrow C E=A B_{1} \\
\Rightarrow \frac{\left|A B_{2}\right|}{\left|B_{2} C_{2}\right|}=\frac{A E}{E B_{1}}=\frac{4}{3} .
\end{array}
$$