Let $f(x)=\frac{4^{x}}{4^{x}+2}$, then the sum
$$
f\left(\frac{1}{1001}\right)+f\left(\frac{2}{1001}\right)
$$
$+f\left(\frac{3}{1001}\right)+\cdots+f\left(\frac{1000}{1001}\right)$ is
The original sum is $f\left(\frac{1}{1986}\right)+f\left(\frac{2}{1986}\right)$
$+\cdots+f\left(\frac{1985}{1986}\right)$. Considering that the number 1986 is non-essential, and given that using year numbers in problems is common, to avoid unnecessary confusion, this number has been modified. This problem is tricky, and brute force calculation is not feasible. Inspired by Gauss's method of summing, one should naturally first calculate $f\left(\frac{1}{1001}\right)+f\left(\frac{1000}{1001}\right)$. Generally, if $x+y=1$, then calculate $f(x)+f(y)=\frac{4^{x}}{4^{x}+2}$
$+\frac{4^{y}}{4^{y}+2}=1$. The sum has 1000 terms, so the answer should be 500. In fact, we can find the sum in the general case:
$$
\begin{array}{l}
f\left(\frac{1}{n}\right)+f\left(\frac{2}{n}\right)+\cdots+f\left(\frac{n-1}{n}\right) \\
=\frac{n-1}{2} .
\end{array}
$$

To standardize the answer, we chose a not too small number $n=1001$, hinting that candidates must "extract" the solution.