Example 4. Solve the system of equations:
$$
\left\{\begin{array}{l}
\left|a_{1}-a_{2}\right| x_{2}+\left|a_{1}-a_{3}\right| x_{3}+\left|a_{1}-a_{4}\right| x_{4}=1, \\
\left|a_{2}-a_{1}\right| x_{1}+\left|a_{2}-a_{3}\right| x_{3}+\left|a_{2}-a_{4}\right| x_{4}=1, \\
\left|a_{3}-a_{1}\right| x_{1}+\left|a_{3}-a_{2}\right| x_{2}+\left|a_{3}-a_{4}\right| x_{4}=1, \\
\left|a_{4}-a_{1}\right| x_{1}+\left|a_{4}-a_{2}\right| x_{2}+\left|a_{4}-a_{3}\right| x_{3}=1 .
\end{array}\right.
$$

where \(a_{1}, a_{2}, a_{3}, a_{4}\) are distinct real numbers.
Without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>a_{4}$, then the original system of equations becomes:
$$
\begin{array}{l}
\left(a_{1}-a_{2}\right) x_{2}+\left(a_{1}-a_{3}\right) x_{3}+\left(a_{1}-a_{4}\right) x_{4}=1, \\
\left(a_{2}-a_{1}\right) x_{1}+\left(a_{2}-a_{3}\right) x_{3}+\left(a_{2}-a_{4}\right) x_{4}=1, \\
\left(a_{3}-a_{1}\right) x_{2}+\left(a_{3}-a_{2}\right) x_{2}+\left(a_{3}-a_{4}\right) x_{4}=1, \\
\left(a_{4}-a_{1}\right) x_{1}+\left(a_{4}-a_{2}\right) x_{2}+\left(a_{4}-a_{3}\right) x_{3}=1 .
\end{array}
$$

Subtracting (2) from (1), (3) from (2), and (4) from (3), we simplify to get:
$$
\left\{\begin{array}{l}
x_{2}+x_{3}+x_{4}-x_{1}=0, \\
x_{1}+x_{2}-x_{3}-x_{4}=0, \\
x_{1}+x_{2}+x_{3}-x_{4}=0 .
\end{array}\right.
$$

Solving these, we get $x_{2}=x_{3}=0, x_{1}=x_{4}$. Then, using (1) and (4), we solve to get
$$
x_{1}=x_{4}=\frac{1}{a_{1}-a_{4}} .
$$