2. Given $m^{2}+71 m-1997=0, n^{2}+71 n$ $-1997=0$, and $m \neq n$. Then $\frac{1}{m}+\frac{1}{n}=$ $\qquad$ .
2. $\frac{71}{1997}$.

It is easy to know that $m$ and $n$ are two distinct roots of the equation $x^{2}+71 x-1997=0$, then
$$
\frac{1}{m}+\frac{1}{n}=\frac{m+n}{m n}=\frac{-71}{-1997}=\frac{71}{1997} .
$$