Given a positive integer $n(n \geq 5)$, try to provide a set of distinct positive even numbers $p_{1}, p_{2}, \ldots, p_{n}$, such that
$$
\frac{1}{p_{1}}+\frac{1}{p_{2}}+\ldots+\frac{1}{p_{n}}=\frac{2003}{2002} .
$$
Solution: Since $\frac{2003}{2002}=1+\frac{1}{2002}$, we can construct some $p_{i}$ such that the sum of their reciprocals is 1.
Because $\frac{1}{k}-\frac{1}{k+1}=\frac{1}{k(k+1)}$, then
$$
\frac{1}{k}=\frac{1}{k+1}+\frac{1}{k(k+1)}.
$$

Using identity 11 to construct the required expression:
$$
\begin{aligned}
1 & =\frac{1}{2}+\frac{1}{2}=\frac{1}{2}+\left(\frac{1}{3}+\frac{1}{2 \times 3}\right) \\
& =\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{1}{2}+\frac{1}{6}+\left(\frac{1}{4}+\frac{1}{3 \times 4}\right) \\
& =\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{12}.
\end{aligned}
$$

When $n=5$, take $p_{1}, p_{2}, \ldots, p_{5}$ as $2, 4, 6, 12, 2002$ respectively.

When $n>5$, for the largest positive even number, continue to use (1) for further processing, for example, $\frac{1}{12}=\frac{1}{2}\left(\frac{1}{7}+\frac{1}{42}\right)=\frac{1}{14}+\frac{1}{84}$, etc., to further expand the set of numbers that satisfy the condition. A rigorous statement is left to the reader. (Zhang Yanwei, Suqian Education Bureau, Jiangsu Province, 223800)