Example 6 Given that $p$, $q$, $\frac{2p-1}{q}$, $\frac{2q-1}{p}$ are all integers, and $p>1$, $q>1$. Try to find the value of $p+q$.
Analysis: The conditions of this problem do not provide specific numbers. Yet, it requires finding the value of $p+q$, which poses a certain difficulty. If we analyze the given conditions one by one, we almost get no useful information. However, if we look at the conditions as a whole, i.e., consider $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ together, we will find that at least one of $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ is less than 2. Otherwise, if $\frac{2 p-1}{q} \geqslant 2$ and $\frac{2 q-1}{p} \geqslant 2$ were both true, then we would have
$2 p-1 \geqslant 2 q, 2 q-1 \geqslant 2 p$.
Adding these two inequalities gives
$$
2 p+2 q-2 \geqslant 2 p+2 q \text {. }
$$

which simplifies to
$$
-2 \geqslant 0 \text{. }
$$
This is clearly impossible.
By observing the given conditions as a whole, we have found a way to solve the problem.

Since at least one of $\frac{2 p-1}{q}$ and $\frac{2 q-1}{p}$ is less than 2, let's assume $\frac{2 p-1}{q} < 2$. From $\frac{2 p-1}{q} < 2$, we get $2 p - 1 < 2 q$, or $2 p - 2 q < 1$. Since $p$ and $q$ are integers, the only possible solution is $2 p - 2 q = 0$, which implies $p = q$. However, this contradicts the condition that $\frac{2 p-1}{q} < 2$. Therefore, we need to consider the next possible integer values. From $\frac{2 p-1}{q} < 2$, we can try $p = 3$ and $q = 5$. Substituting these values, we get $\frac{2 \cdot 3 - 1}{5} = \frac{5}{5} = 1 < 2$, which satisfies the condition. Thus, $p = 3$ and $q = 5$. Therefore, $p+q=8$.