4・142 Solve the system of equations
$$\left\{\begin{array}{l}
x_{1} \cdot x_{2} \cdot x_{3}=x_{1}+x_{2}+x_{3}, \\
x_{2} \cdot x_{3} \cdot x_{4}=x_{2}+x_{3}+x_{4}, \\
x_{3} \cdot x_{4} \cdot x_{5}=x_{3}+x_{4}+x_{5}, \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
x_{1985} \cdot x_{1986} \cdot x_{1987}=x_{1985}+x_{1986}+x_{1987}, \\
x_{1986} \cdot x_{1987} \cdot x_{1}=x_{1986}+x_{1987}+x_{1}, \\
x_{1987} \cdot x_{1} \cdot x_{2}=x_{1987}+x_{1}+x_{2} .
\end{array}\right.$$
[Solution] (1) - (2), we get $x_{2} \cdot x_{3}\left(x_{1}-x_{4}\right)=x_{1}-x_{4}$. Therefore, $x_{2} x_{3}=1$ or $x_{1}=x_{4}$.

If $x_{2} \cdot x_{3}=1$, then from (1) we get $x_{1}=x_{1}+x_{2}+x_{3}$, thus $x_{2}+x_{3}=0$, but the system of equations $x_{2} \cdot x_{3}=1$ and $x_{2}+x_{3}=0$ has no real solutions. Therefore, it must be that $x_{1}=x_{4}$.

Similarly, we can get $x_{2}=x_{5}, x_{3}=x_{6}, \cdots, x_{1985}=x_{1}, x_{1986}=x_{2}$, thus $x_{3}=$ $x_{6}=x_{9}=\cdots=x_{1986}=x_{2}=x_{5}=x_{8}=\cdots=x_{1985}=x_{1}=x_{4}=x_{7}$ $=\cdots=x_{1984}=x_{1987}$. Therefore, the solution to the original system of equations is of the form $(x, x, \cdots, x)$ where $x$ is some real number. Substituting into equation (1) gives $x^{3}=3 x$, solving which yields $x=0$, or $x= \pm \sqrt{3}$.