18. 2.11 $\stackrel{\star}{\star}$ Determine whether the following system of equations has positive integer solutions:
$$
\left\{\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{1985}^{2}=y^{3}, \\
x_{1}^{3}+x_{2}^{3}+\cdots+x_{1985}^{3}=z^{2},
\end{array}\right.
$$

where $x_{i} \neq x_{j} (i \neq j)$.
For any positive integer $n$, the system of equations
$$
\left\{\begin{array}{l}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=y^{3} \\
x_{1}^{3}+x_{2}^{3}+\cdots+x_{n}^{3}=z^{2}
\end{array}\right.
$$

has infinitely many positive integer solutions. In fact, let $x_{1}=k, x_{2}=2 k, \cdots, x_{n}=n k$, substituting into (1), (2) we get
$$
y^{3}=\frac{1}{6} k^{2} n(n+1)(2 n+1), z^{2}=k^{3}\left(\frac{1}{2} n(n+1)\right)^{2} .
$$

It is only necessary to take $k=\left(\frac{1}{6} n(n+1)(2 n+1)\right)^{m}$, where $m \equiv 4(\bmod 6)$ to ensure that $y$ and $z$ are positive integers.

Taking $n=1985, m=4$, we get a positive integer solution to the original system of equations: $x_{1}=k, x_{2}=2 k, \cdots, x_{1985}=1985 \cdot k$, where $k=\left(\frac{1}{6} \times 1985 \times 1986 \times 3971\right)^{4}$.