一、(Full marks 50 points) (1) Find natural numbers $x, y$ such that $\frac{6}{1997}=\frac{1}{x}+\frac{1}{y}$.
(2) How many natural numbers $n(n<1997)$ can make $\frac{n}{1097}=\frac{1}{x}+\frac{1}{y} ?(x, y$ are natural numbers $)$
(1) From $\frac{6}{1997}=\frac{x+y}{x y}$, we have $6 x y=1997(x+y)$.
Since $1997$ is a prime number,
$\therefore 6 \mid x+y$.
Let $\left\{\begin{array}{l}x+y=6 k^{2}, \\ x y=1997 k^{2}\end{array}\right.$.
To find $k$, we can take $x=k, y=1997 k$
Substituting into (1) gives $k+1997 k=6 k^{2}, 1998=6 k$.
Thus, $k=333$.
$$
\therefore x=333, y=1997 \times 333 \text {. }
$$
$\frac{6}{1997}=\frac{1}{333}+\frac{1}{1997 \times 333}$ is the solution.
(2) From $\frac{n}{1997}=\frac{x+y}{x y}$, we have $n x y=1997(x+y)$.

Taking $x=k, y=1997 k$ and substituting gives $x+y=n k^{2}$.
Thus, $k+1997 k=n k^{2}, 1998=n k$.
We know that $n$ is a divisor of 1998.
$$
\because 1998=2 \times 3^{3} \times 37 \text {, }
$$

The number of divisors $d=(1+1)(3+1)(1+1)=16$. v can make $\frac{n}{1997}=\frac{1}{x}+\frac{1}{y}$.