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 subscripts $i$ and $j$, the original system remains unchanged. Therefore, without loss of generality, we can assume $a_{1}>a_{2}>a_{3}>a_{4}$. At this point, the original system of equations 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
$$
x_{j}=x_{k}=0, x_{i}=x_{l}=\frac{1}{a_{i}-a_{l}} .
$$