4. The real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=a_{1} ; \\
x_{2}+x_{3}+x_{1}=a_{2} ; \\
x_{3}+x_{4}+x_{5}=a_{3} ; \\
x_{4}+x_{5}+x_{1}=a_{4} ; \\
x_{5}+x_{1}+x_{2}=a_{5} .
\end{array}\right.
$$

where $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are real constants, and $a_{1}>a_{2}>a_{3}>a_{4}>a_{5}$. Then the order of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ is
(A) $x_{1}>x_{2}>x_{3}>x_{4}>x_{5}$;
(B) $x_{4}>x_{2}>x_{1}>x_{3}>x_{5}$;
(C) $x_{3}>x_{1}>x_{4}>x_{2}>x_{5}$;
(D) $x_{5}>x_{3}>x_{1}>x_{4}>x_{2}$.
4. (C)

Given the equations in the system, subtracting them in sequence pairwise yields
$$
\begin{array}{ll}
x_{1}-x_{4}=a_{1}-a_{2}, & x_{2}-x_{5}=a_{2}-a_{3}, \\
x_{3}-x_{1}=a_{3}-a_{4}, & x_{4}-x_{2}=a_{4}-a_{5} . \\
\because \quad a_{1}>a_{2}>a_{3}>a_{4}>a_{5}, & \\
\therefore \quad x_{1}>x_{4}, x_{2}>x_{5}, x_{3}>x_{1}, x_{4}>x_{2} .
\end{array}
$$

Thus, we have $x_{3}>x_{1}>x_{4}>x_{2}>x_{5}$.