Solve the following system of equations:

$$
\begin{aligned}
& \left|a_{1}-a_{2}\right| \cdot x_{2}+\left|a_{1}-a_{3}\right| \cdot x_{3}+\left|a_{1}-a_{4}\right| \cdot x_{4}=1 \\
& \left|a_{2}-a_{1}\right| \cdot x_{1} \quad+\left|a_{2}-a_{3}\right| \cdot x_{3}+\left|a_{2}-a_{4}\right| \cdot x_{4}=1 \\
& \left|a_{3}-a_{1}\right| \cdot x_{1}+\left|a_{3}-a_{2}\right| \cdot x_{2} \quad+\left|a_{3}-a_{4}\right| \cdot x_{4}=1 \\
& \left|a_{4}-a_{1}\right| \cdot x_{1}+\left|a_{4}-a_{2}\right| \cdot x_{2}+\left|a_{4}-a_{2}\right| \cdot x_{3} \quad=1
\end{aligned}
$$

where $a_{1}, a_{2}, a_{3}$, and $a_{4}$ represent four different given real numbers.
) First, we solve the system for the case of the parameter size relationship $a_{1}<a_{2}<a_{3}<a_{4}$. We will reduce other cases to this one. (I) is thus structured:

$$
\begin{array}{ccc}
\quad\left(a_{2}-a_{1}\right) x_{2} & +\left(a_{3}-a_{1}\right) x_{3}+\left(a_{4}-a_{1}\right) x_{4} & =1 \\
\left(a_{2}-a_{1}\right) x_{1} & +\left(a_{3}-a_{2}\right) x_{3}+\left(a_{4}-a_{2}\right) x_{4} & =1 \\
\left(a_{3}-a_{1}\right) x_{1}+\left(a_{3}-a_{2}\right) x_{2} & +\left(a_{4}-a_{3}\right) x_{4} & =1 \\
\left(a_{4}-a_{1}\right) x_{1}+\left(a_{4}-a_{2}\right) x_{2} & +\left(a_{4}-a_{3}\right) x_{3} & =1
\end{array}
$$

Subtract (II)-(IV) from the subsequent equations, and add the first and the last equation. With appropriate factoring:

$$
\begin{aligned}
& \left(a_{2}-a_{1}\right)\left(x_{1}-x_{2}-x_{3}-x_{4}\right)=0 \\
& \left(a_{3}-a_{2}\right)\left(x_{1}+x_{2}-x_{3}-x_{4}\right)=0 \\
& \left(a_{4}-a_{3}\right)\left(x_{1}+x_{2}+x_{3}-x_{4}\right)=0 \\
& \left(a_{4}-a_{1}\right)\left(x_{1}+x_{2}+x_{3}+x_{4}\right)=2
\end{aligned}
$$

Since the highlighted differences of the parameters are not zero in any equation, we can simplify with them:

$$
\begin{aligned}
& x_{1}-x_{2}-x_{3}-x_{4}=0 \\
& x_{1}+x_{2}-x_{3}-x_{4}=0 \\
& x_{1}+x_{2}+x_{3}-x_{4}=0 \\
& x_{1}+x_{2}+x_{3}+x_{4}=\frac{2}{a_{4}-a_{1}}
\end{aligned}
$$

Now, from the difference of (VII) and (VI), as well as (VIII) and (VII), and the difference of (IX) and (VIII), and the sum of (VI) and (IX):

$$
x_{2}=x_{3}=0, \quad x_{4}=x_{1}=\frac{1}{a_{4}-a_{1}}
$$

$\beta$ ) Let $a_{h}<a_{i}<a_{j}<a_{k}$, where $h, i, j, k$ are any permutation of the indices 1, 2, 3, 4. The anticipated reduction is carried out with the following substitution. Let

$$
\begin{array}{llll}
a_{h}=b_{1}, & a_{i}=b_{2}, & a_{j}=b_{3}, & a_{k}=b_{4} \\
x_{h}=y_{1}, & x_{i}=y_{2}, & x_{j}=y_{3}, & x_{k}=y_{4}
\end{array}
$$

The $h$-th equation of (I), disregarding the order of terms on the left side, can be written as:

$$
\left|a_{h}-a_{i}\right| \cdot x_{i}+\left|a_{h}-a_{j}\right| \cdot x_{j}+\left|a_{h}-a_{k}\right| \cdot x_{k}=1
$$

Therefore, it is replaced by:

$$
\left|b_{1}-b_{2}\right| \cdot y_{2}+\left|b_{1}-b_{3}\right| \cdot y_{3}+\left|b_{1}-b_{4}\right| \cdot y_{4}=1
$$

This differs from (II) only in that every $a$ and $x$ is replaced by $b$ and $y$, respectively, since $b_{1}<b_{2}<b_{3}<b_{4}$, each coefficient is equal to the negative of the difference inside the absolute value. Similarly, the $i$-th, $j$-th, and $k$-th equations of (I) are replaced by (III), (IV), and (V), respectively, with the mentioned substitution of letters, while the indices remain unchanged everywhere.

Accordingly, for the case $a_{h}<a_{i}<a_{j}<a_{k}$, the solution of (I) is:

$$
x_{h}=x_{k}=\frac{1}{a_{k}-a_{h}}, \quad x_{i}=x_{j}=0
$$

József Berács (Győr, Czuczor G. Bencés Gymnasium, 3rd grade)