Example 4 Given real numbers $x, y, z, u$ satisfy
$$
\frac{x}{y+z+u}=\frac{y}{z+u+x}=\frac{z}{u+x+y}=\frac{u}{x+y+z} \text {. }
$$

Find the value of $\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z}$.
Let
$$
\frac{x}{y+z+u}=\frac{y}{z+u+x}=\frac{z}{u+x+y}=\frac{u}{x+y+z}=k \text {. }
$$

Then $x=k(y+z+u), y=k(z+u+x)$,
$$
z=k(u+x+y), u=k(x+y+z) \text {. }
$$

Adding the above four equations yields
$$
x+y+z+u=3 k(x+y+z+u) \text {. }
$$
(1) If $x+y+z+u=0$, then
$$
\begin{array}{l}
\frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z} \\
=-1+(-1)+(-1)+(-1)=-4 .
\end{array}
$$
(2) If $x+y+z+u \neq 0$, then
$$
\begin{array}{l}
k=\frac{1}{3} \Rightarrow x=\frac{1}{3}(y+z+u) \\
\Rightarrow x=y=z=u=\frac{1}{4}(x+y+z+u) \\
\Rightarrow \frac{x+y}{z+u}+\frac{y+z}{u+x}+\frac{z+u}{x+y}+\frac{u+x}{y+z}=4 .
\end{array}
$$