4・150 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 known distinct real numbers.
[Solution] We notice that in the system of equations, if we swap the subscript $i$ with $j$, and $j$ with the equations remain unchanged. Therefore, without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>a_{4}$. The system of equations then becomes:
$$\left\{\begin{array}{l}
\left(a_{1}-a_{2}\right) x_{2}+\left(a_{1}-a_{3}\right) x_{3}+\left(a_{1}-a_{4}\right) x_{4}=1, \\
\left(a_{1}-a_{2}\right) x_{1}+\left(a_{2}-a_{3}\right) x_{3}+\left(a_{2}-a_{4}\right) x_{4}=1, \\
\left(a_{1}-a_{3}\right) x_{2}+\left(a_{2}-a_{3}\right) x_{2}+\left(a_{3}-a_{4}\right) x_{4}=1, \\
\left(a_{1}-a_{4}\right) x_{1}+\left(a_{2}-a_{4}\right) x_{2}+\left(a_{3}-a_{4}\right) x_{3}=1
\end{array}\right.$$
Subtracting (2) from (1), (3) from (2), and (4) from (3), we get
$$\left\{\begin{array}{l}
\left(a_{1}-a_{2}\right)\left(x_{2}+x_{3}+x_{4}-x_{1}\right)=0, \\
\left(a_{2}-a_{3}\right)\left(-x_{1}-x_{2}+x_{3}+x_{4}\right)=0, \\
\left(a_{3}-a_{4}\right)\left(-x_{1}-x_{2}-x_{3}+x_{4}\right)=0 .
\end{array}\right.$$

Since $a_{1}-a_{2} \neq 0, a_{2}-a_{3} \neq 0, a_{3}-a_{4} \neq 0$, we have
$$\left\{\begin{array}{l}
x_{2}+x_{3}+x_{4}-x_{1}=0 \\
-x_{1}-x_{2}+x_{3}+x_{4}=0 \\
-x_{1}-x_{2}-x_{3}+x_{4}=0
\end{array}\right.$$
Subtracting (6) from (5), and (7) from (6), we get
$$x_{2}=0, x_{3}=0 .$$

Substituting into (5) gives $x_{1}=x_{4}$.
Substituting into (1) and (4) gives $x_{1}=x_{4}=\frac{1}{a_{1}-a_{4}}$.
Upon verification, when $a_{1}>a_{2}>a_{3}>a_{4}$,
$$x_{1}=x_{3}=0, x_{1}=x_{4}=\frac{1}{a_{1}-a_{4}}$$

is a solution to the system of equations.
In general, when $a_{i}>a_{j}>a_{k}>a_{l}$, the solution to the system of equations is
$$x_{j}=x_{k}=0, x_{i}=x_{l}=\frac{1}{a_{i}-a_{l}} .$$