Solution(from the Book "International Mathematical Olympiads" by "Greitzer Samuel, Klamkin Murrays, John Scholes"): The $ i$'th equation can be written in the form: $ \sum^{4}_{j = 1}\left|a_{i}-a_{j}\right|x_{j}= 1,\;\;\;i = 1,2,3,4$ the solutions of this system are unchangeable under any permutation of indices $ 1,2,3,4$! In other words if $ x_{1},\cdots,x_{4}$ is a solution of this system by a defined selection of $ a_{1},\cdots,a_{4}$, and then we permute $ a_{i}$'s, in this case the same permutation for $ x_{1},\cdots,x_{4}$ will be a solution of the system. because $ a_{i}$'s are distinct let: $ a_{1}> a_{2}> a_{3}> a_{4}$ lets write the equations in the following form: $ \begin{array}{c}\;0+\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\;\;\;\;\;\textbf{(1)}\ \ \left(a_{1}-a_{2}\right)x_{1}+0+\left(a_{2}-a_{3}\right)x_{3}+\left(a_{2}-a_{4}\right)x_{4}= 1\;\;\;\;\;\textbf{(2)}\ \ \left(a_{1}-a_{3}\right)x_{1}+\left(a_{2}-a_{3}\right)x_{2}+0+\left(a_{3}-a_{4}\right)x_{4}= 1\;\;\;\;\;\textbf{(3)}\ \ \left(a_{1}-a_{4}\right)x_{1}+\left(a_{2}-a_{4}\right)x_{2}+\left(a_{3}-a_{4}\right)x_{3}+0 = 1\;\;\;\;\;\textbf{(4)}\end{array}$ now lets subtract each equation from its next equation: as a consequence $ \textbf{(2)-(1)}$(subtract $ \textbf{(1)}$ from $ \textbf{(2)}$) gives: $ \left(a_{1}-a_{2}\right)\left(x_{1}-x_{2}-x_{3}-x_{4}\right) = 0$ and because $ a_{1}-a_{2}\neq0$ by dividing by $ a_{1}-a_{2}$: $ x_{1}-x_{2}-x_{3}-x_{4}= 0\;\;\;\;\;\textbf{(5)}$ by the same way, $ \textbf{(3)-(2)}$ gives: $ x_{1}+x_{2}-x_{3}-x_{4}= 0\;\;\;\;\;\textbf{(6)}$ $ \textbf{(4)-(3)}$ gives: $ x_{1}+x_{2}+x_{3}-x_{4}= 0\;\;\;\;\;\textbf{(7)}$ Now $ \textbf{(6)-(5)}$ gives $ x_{2}= 0$ and $ \textbf{(7)-(6)}$ gives $ x_{3}= 0$ and atlast by $ \textbf{(5),(6),(7)}$ we have $ x_{4}= x_{1}$ by putting these results in four equations of the system we have: $ \boxed{x_{1}= x_{4}=\frac{1}{a_{1}-a_{4}},\;\;\;\;\;x_{2}= x_{3}= 0}$

My Solution: $\begin{cases} & \text{ } |a_{1}-a_{2}|x_{2}+|a_{1}-a_{3}|x_{3}+|a_{1}-a_{4}|x_{4}=1 \\ & \text{ } |a_{2}-a_{1}|x_{1}+|a_{2}-a_{3}|x_{3}+|a_{2}-a_{4}|x_{4}=1 \\ & \text{ } |a_{3}-a_{1}|x_{1}+|a_{3}-a_{2}|x_{2}+|a_{3}-a_{4}|x_{4}=1 \\ & \text{ } |a_{4}-a_{1}|x_{1}+|a_{4}-a_{2}|x_{2}+|a_{4}-a_{3}|x_{3}=1 \\ \end{cases}.$ + Let :$ |a_{1}-a_{2}|=a;|a_{1}-a_{3}|=b;|a_{1}-a_{4}|=c; |a_{2}-a_{3}|=d;|a_{2}-a_{4}|=e;|a_{3}-a_{4}|=f;$ We have the system of equations: $\begin{cases} & \text{ } a x_2 + b x_3 + c x_4 =1; (1) \\ & \text{ } a x_1 + d x_3+e x_4 =1; (2) \\ & \text{ } b x_1 + d x_2 + f x_4 =1; (3) \\ & \text{ } c x_1 + e x_2+f x_3 =1; (4) \\ \end{cases};$ Changes to the following: *): $(1) \times e - (2) \times c \Rightarrow $: $-ac x_1 + ae x_2 + (be-dc)x_3 = e-c$; $(1')$ *): $(1) \times f - (3) \times c \Rightarrow $: $-cbx_1 + (af-dc)x_2 + bfx_3 = f-c$ ; $ (2')$ and $ c x_1 + e x_2+f x_3 =1; (4)$ Combining equations $(1')$; $(2')$ and $(4)$, We have a new system of equations, only $x_1; x_2; x_3$ At this point, we can solve the $3$ equations with three vars, respectively, Using a similar way as above. Results: $\boxed{\left (x_1;x_2;x_3;x_4 \right )=(\frac{1}{a_{1}-a_{4}};0;0;\frac{1}{a_{1}-a_{4}})}$