A13 (8-5, Czechoslovakia) 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.
If we swap the indices of $a_{i}, a_{j}$, the original system of equations remains unchanged. Therefore, we can assume $a_{1}>a_{2}>a_{3}>a_{4}$. Thus, the original system of equations becomes:
$$
\left\{\begin{array}{l}
\left(a_{1}-a_{2}\right) x_{2}+\left(a_{1}-a_{3}\right) x_{3}+\left(a_{1}-a_{4}\right) x_{4}=1, \\
\left(a_{1}-a_{2}\right) x_{1}+\left(a_{2}-a_{3}\right) x_{3}+\left(a_{2}-a_{4}\right) x_{4}=1, \\
\left(a_{1}-a_{3}\right) x_{1}+\left(a_{2}-a_{3}\right) x_{2}+\left(a_{3}-a_{4}\right) x_{4}=1, \\
\left(a_{1}-a_{4}\right) x_{1}+\left(a_{2}-a_{4}\right) x_{2}+\left(a_{3}-a_{4}\right) x_{3}=1 .
\end{array}\right.
$$
(1) - (2), we get
$$
\left(a_{1}-a_{2}\right)\left(x_{3}+x_{2}+x_{4}-x_{1}\right)=0 ;
$$
(2) - (3), we get
$$
\left(a_{2}-a_{3}\right)\left(x_{3}+x_{4}-x_{1}-x_{2}\right)=0 ;
$$
(3) - (4), we get
$$
\left(a_{3}-a_{4}\right)\left(x_{4}-x_{1}-x_{2}-x_{3}\right)=0 .
$$

Since $a_{i}$ are all distinct, we have
$$
\left\{\begin{array}{l}
x_{2}+x_{3}+x_{4}-x_{1}=0, \\
x_{3}+x_{4}-x_{1}-x_{2}=0, \\
x_{4}-x_{1}-x_{2}-x_{3}=0 .
\end{array}\right.
$$
(5) - (6), we get
$$
\begin{array}{l}
x_{2}=0 ; \\
x_{3}=0 .
\end{array}
$$
(6) - (7), we get

And $x_{1}$ only needs to equal $x_{4}$ to satisfy (5), (6), and (7).
Substituting back into (1), we get the unique solution
$$
x_{1}=x_{4}=\frac{1}{a_{1}-a_{4}}, x_{2}=x_{3}=0 .
$$