Example 8 When $x$ equals $\frac{1}{1993}, \frac{1}{1992}, \ldots, \frac{1}{2}, 1, 2$, $\cdots, 1992, 1993$, calculate the value of the algebraic expression $\frac{x^{2}}{1+x^{2}}$, then add up all the results obtained. What is the sum?
(1993-1994 Wuhan and Four Cities Mathematics Competition)
Solution: Let $A(x)=\frac{x^{2}}{1+x_{1}^{2}}$, by substituting $x$ with $\frac{1}{x}$, we get the conjugate of $A(x)$, denoted as $B(x)$, which is $B(x)=A\left(\frac{1}{x}\right)=$ $\frac{1}{1+x^{2}}$. Then,
$$
A(x)+B(x)=1 .
$$

Thus, the desired sum
$$
\begin{aligned}
S= & A\left(\frac{1}{1993}\right)+\cdots+A\left(\frac{1}{2}\right)+A(1) \\
& +\cdots+A(1993) \\
= & B(1993)+\cdots+B(2)+A(1)+\cdots \\
& +A(1993) \\
= & {[B(1993)+A(1993)]+[B(1992)} \\
& +A(1992)]+\cdots+[(B(2) \\
& +A(2)]+A(1)=1992.5 .
\end{aligned}
$$