254. Solve the system of four equations with four unknowns

$$
\begin{array}{rr}
|a-b| y+|a-c| z+|a-d| t & =1 \\
|b-a| x+|b-c| z+|b-d| t & =1 \\
|c-a| x+|c-b| y+|c-d| t & =1 \\
|d-a| x+|d-b| y+|d-c| z & =1
\end{array}
$$

where $a, b, c, d$ are some four pairwise distinct real numbers.
254. To emphasize the complete symmetry of the equations of our system with respect to the unknowns and the coefficients of these unknowns, let us denote $x=x_{1}, y=x_{2}, z=x_{3}$ and $t=x_{4}; a=a_{1}, b=a_{2}, c=a_{3}, d=a_{4}$; in this case, our equations can be written as:

$$
\sum_{j=1}^{4}\left|a_{i}-a_{j}\right| x_{j}=1 \quad(i=1,2,3,4)
$$

Further, let, for example, $a_{1}>a_{2}>a_{3}>a_{4}$; then we have

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

Subtract the second equation of the system (**) from the first, the third from the second, and the fourth from the third; we obtain:

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

Since the numbers $a_{1}, a_{2}, a_{3}, a_{4}$ are pairwise distinct, then

$$
x_{1}=x_{2}+x_{3}+x_{4}, \quad x_{1}+x_{2}=x_{3}+x_{4}, \quad x_{1}+x_{2}+x_{3}=x_{4}
$$

from which it follows that

$$
x_{2}=x_{3}=0 . \quad x_{1}=x_{4}
$$

and then (for example, from the last of the equations (**)) we find:

$$
x_{1}=\frac{1}{a_{1}-a_{4}}
$$

Verification shows that the values $x_{2}=x_{3}=0, x_{1}=x_{4}=\frac{1}{a_{1}-a_{4}}$ indeed satisfy all the equations of the system.

Answer: if $a>b>c>d$, then $\quad x=t=\frac{1}{a-d}, y=z=0$.