20. (ROM 3) Solve the system of equations
$$
\begin{array}{c}
x_{1}\left|x_{1}\right|=x_{2}\left|x_{2}\right|+\left(x_{1}-a\right)\left|x_{1}-a\right|, \\
x_{2}\left|x_{2}\right|=x_{3}\left|x_{3}\right|+\left(x_{2}-a\right)\left|x_{2}-a\right|, \\
\cdots \\
x_{n}\left|x_{n}\right|=x_{1}\left|x_{1}\right|+\left(x_{n}-a\right)\left|x_{n}-a\right|,
\end{array}
$$
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 $\alpha$, then $\left(-x_{1},-x_{2}\right.$, $\left.\ldots,-x_{n}\right)$ satisfies the system with parameter $-a$. Hence it is sufficient to consider only $a \geq 0$.
Let $\left(x_{1}, \ldots, x_{m}\right)$ 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.