6. Let $p, q, \frac{2 p-1}{q}, \frac{2 q-1}{p}$ all be integers, and $p>1, q>$
1. Then $p+q=$ $\qquad$ .
6. $p+q=8$.

From $\frac{2 q-1}{p}, \frac{2 p-1}{q}$ both being positive integers, we know that one of them must be less than 2. Otherwise, $\frac{2 q-1}{p}>2, \frac{2 p-1}{q}>2$, leading to $2 p-1+2 q-1>2 p+2 q$, which is a contradiction. Assume $0<\frac{2 q-1}{p}<2$, then it must be that $2 q-1=p$. It is easy to see that $q=3, p=5$, thus $p+q=8$.

Note: The original problem statement has a typo in the final result, which should be $p+q=8$ instead of $p+q=3$.