Example 30 (2003 Beijing Competition Question) Given the function $f(x)=\frac{2+x}{1+x}$. Let $f(1)+f(2)+\cdots+f(1000)=m$, $f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)+\cdots+f\left(\frac{1}{1000}\right)=n$, then the value of $m+n$ is $\qquad$ .
Solve: Fill in 2998.5. Reason: The symmetric form of the arguments in the given problem, such as $f(2)$ and $f\left(\frac{1}{2}\right), \cdots, f(1000)$ and $f\left(\frac{1}{1000}\right)$, suggests combining them for solving. It is easy to see that $x \neq -1$, and since
$$
f(x)+f\left(\frac{1}{x}\right)=\frac{2+x}{1+x}+\frac{2+\frac{1}{x}}{1+\frac{1}{x}}=\frac{(2+x)+(2 x+1)}{1+x}=3,
$$

and $f(1)=\frac{2+1}{1+1}=\frac{3}{2}$, thus
$$
\begin{aligned}
m+n & =[f(1)+f(2)+\cdots+f(1000)]\left[f\left(\frac{1}{2}\right)+f\left(\frac{1}{3}\right)+\cdots+f\left(\frac{1}{1000}\right)\right] \\
& =f(1)+999 \cdot\left[f(x)+f\left(\frac{1}{x}\right)\right] \\
& =\frac{3}{2}+2997=2998.5
\end{aligned}
$$