If $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ are all integers, and $p>1$, $q>1$, find the value of $p+q$.
Solution 1 First, $p \neq q$.
Indeed, if $p=q$, then
\[
\frac{2 p-1}{q}=\frac{2 p-1}{p}=2-\frac{1}{p} \quad \text{, since } p>1, \quad \frac{2 p-1}{q} \text{ is not an integer, contradicting the problem statement.}
\]

By symmetry, assume without loss of generality that $p>q$, and let $\frac{2 q-1}{p}=m$, then $m$ is a positive integer.
\[
\because \quad m p=2 q-1, \quad \therefore \quad q=3.
\]
Thus, $p=2 q-1=5$.
\[
\therefore \quad p+q=8 \text{.}
\]

Solution 2 From Solution 1, we know $p \neq q$, assume without loss of generality that $p>q$,
Let $\frac{2 p-1}{q}=m$,
\[
\frac{2 q-1}{p}=n \text{.}
\]

Then $m, n$ are positive integers, and it is easy to see that $m>n$.
From (2), $q=\frac{n p+1}{2}$,
Substituting (3) into (1), we get
\[
\begin{array}{l}
2 p-1=m q=m \cdot \frac{n p+1}{2} . \\
\therefore \quad(4-m n) p=m+2,
\end{array}
\]
\[
\therefore \quad 4-m n \text{ is a positive integer.}
\]
That is, $m n=1, m n=2$ or $m n=3$.
Noting that $m>n$, we have only
\[
\left\{\begin{array}{l}
m=2 \\
n=1
\end{array} \text { or } \left\{\begin{array}{l}
m=3 \\
n=1 .
\end{array}\right.\right.
\]

When $m=2, n=1$, from (1) and (2) we get $p=2, q=\frac{3}{2}$, which does not meet the problem's conditions.
When $m=3, n=1$, from (1) and (2) we get
\[
\begin{array}{c}
\quad p=5, q=3 . \\
\therefore \quad p+q=8 .
\end{array}
\]