20. (ROM 3) Solve the system of equations  $$ \begin{gathered} x_{1}\left|x_{1}\right|=x_{2}\left|x_{2}\right|+\left(x_{1}-a\right)\left|x_{1}-a\right| \\ x_{2}\left|x_{2}\right|=x_{3}\left|x_{3}\right|+\left(x_{2}-a\right)\left|x_{2}-a\right| \\ \quad \cdots \\ x_{n}\left|x_{n}\right|=x_{1}\left|x_{1}\right|+\left(x_{n}-a\right)\left|x_{n}-a\right| \end{gathered} $$  in the set of real numbers, where \( a > 0 \).
20. If $\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ satisfies the system with parameter $a$, then $\left(-x_{1},-x_{2}, \ldots,-x_{n}\right)$ satisfies the system with parameter $-a$. Hence it is sufficient to consider only $a \geq 0$. Let $\left(x_{1}, \ldots, x_{n}\right)$ be a solution. Suppose $x_{1} \leq a, x_{2} \leq a, \ldots, x_{n} \leq a$. Summing the equations we get 
$$ \left(x_{1}-a\right)^{2}+\cdots+\left(x_{n}-a\right)^{2}=0 $$
and see that $(a, a, \ldots, a)$ is the only such solution. Now suppose that $x_{k} \geq a$ for some $k$. According to the $k$th equation, 
$$ x_{k+1}\left|x_{k+1}\right|=x_{k}^{2}-\left(x_{k}-a\right)^{2}=a\left(2 x_{k}-a\right) \geq a^{2} $$
which implies that $x_{k+1} \geq a$ as well (here $x_{n+1}=x_{1}$). Consequently, all $x_{1}, x_{2}, \ldots, x_{n}$ are greater than or equal to $a$, and as above $(a, a, \ldots, a)$ is the only solution.