4. Let non-negative real numbers $x_{1}, x_{2}, \cdots, x_{6}$ satisfy
$$
\begin{array}{c}
x_{1}+x_{2}+\cdots+x_{6}=1, x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6} \geqslant \frac{1}{540} . \\
\text { If } \max \left\{x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+x_{4} x_{5} x_{6}+\right. \\
\left.x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2}\right\}=\frac{p}{q},(p, q)=11 \text {, find } p+q .{ }^{[3]}
\end{array}
$$
Let $r=x_{1} x_{3} x_{5}+x_{2} x_{4} x_{6}$,
$$
\begin{aligned}
s= & x_{1} x_{2} x_{3}+x_{2} x_{3} x_{4}+x_{3} x_{4} x_{5}+ \\
& x_{4} x_{5} x_{6}+x_{5} x_{6} x_{1}+x_{6} x_{1} x_{2} .
\end{aligned}
$$

By the AM-GM inequality,
$$
\begin{array}{l}
r+s=\left(x_{1}+x_{4}\right)\left(x_{2}+x_{5}\right)\left(x_{3}+x_{6}\right) \\
\leqslant\left[\frac{\left(x_{1}+x_{4}\right)+\left(x_{2}+x_{5}\right)+\left(x_{3}+x_{6}\right)}{3}\right]^{3}=\frac{1}{27},
\end{array}
$$

with equality if and only if $x_{1}+x_{4}=x_{2}+x_{5}=x_{3}+x_{6}=\frac{1}{3}$. 
Therefore, $s \leqslant \frac{1}{27}-\frac{1}{540}=\frac{19}{540}$.
Let $x_{1}=x_{3}=\frac{3}{10}, x_{5}=\frac{1}{60}, x_{2}=\frac{1}{3}-x_{5}=\frac{19}{60}$,
$$
x_{4}=\frac{1}{3}-x_{1}=\frac{1}{30}, x_{6}=\frac{1}{3}-x_{3}=\frac{1}{30} \text {. }
$$

Then $r=\frac{1}{540}, s=\frac{19}{540}$.
Thus, $p+q=559$.