13. Given that $p$, $q$, $\frac{2 q-1}{p}$, and $\frac{2 p-1}{q}$ are all integers, and $p>1$, $q>1$. Find the value of $p+q$.
13. If $\frac{2 q-1}{p} \geqslant 2, \frac{2 p-1}{q} \geqslant 2$, then
$$
2 q-1 \geqslant 2 p, 2 p-1 \geqslant 2 q \text {. }
$$

Adding the two inequalities gives $2 p+2 q-2 \geqslant 2 p+2 q$. This is clearly a contradiction.
Therefore, at least one of $\frac{2 q-1}{p}$ and $\frac{2 p-1}{q}$ must be less than 2. Suppose $\frac{2 q-1}{p}<2, q>1$, then $\frac{2 q-1}{p}=1$, i.e., $2 q-1=p$.
Also, $\frac{2 p-1}{q}=\frac{4 q-3}{q}$ is an integer, i.e., $4-\frac{3}{q}$ is an integer, so $q=1$ or $q=3$.
Since $q>1$, then $q=3$, and simultaneously $p=2 q-1=5$, hence $p+q=8$.