1. The sum of all irreducible proper fractions with a denominator of 1994 is ( ).
(A) 497.5
(B) 498
(C) 498.5
(D) 997
$-1 . B$.
From $1994=2 \times 997$, and knowing that 997 is a prime number, the sum of the reduced proper fractions in the problem is
$$
\begin{aligned}
N= & \frac{1}{1994}+\frac{3}{1994}+\cdots+\frac{995}{1994}+\frac{999}{1994}+\cdots \\
& +\frac{1993}{1994} \\
= & \frac{1+1993}{1994}+\frac{3+1991}{1994}+\cdots+\frac{995+999}{1994} \\
= & 1+1+\cdots+1 \quad\left(\frac{995+1}{2}=498 \uparrow 1\right) \\
= & 498 .
\end{aligned}
$$