3. If $p$, $q$, $\frac{2 p-1}{q}$, $\frac{2 q-1}{p}$ are all integers, and $p>1$, $q>1$. Then $p+q=$ $\qquad$ .
3.8 .

If $p=q$, then
$$
\frac{2 p-1}{q}=\frac{2 p-1}{p}=2-\frac{1}{p} \text {. }
$$

Given $p>1$, then $\frac{2 p-1}{q}$ is not an integer, which contradicts the problem statement. Therefore, $p \neq q$.
By symmetry, without loss of generality, assume $p>q$. Let
$$
\begin{array}{l}
\frac{2 p-1}{q}=m, \\
\frac{2 q-1}{p}=n .
\end{array}
$$

Then $m$ and $n$ are positive integers, and it is easy to see that $m>n$.
From equation (2), we get
$$
q=\frac{n p+1}{2} \text {. }
$$

Substituting equation (3) into equation (1), we get
$$
2 p-1=m q=m \cdot \frac{n p+1}{2} \text {. }
$$

Simplifying, we get $(4-m n) p=m+2$.
Therefore, $4-m n$ is a positive integer.
Thus, $m n=1$ or 2 or 3.
Since $m>n$, we have $m=2, n=1$ or $m=3, n=1$.
When $m=2, n=1$, from equations (1) and (2), we solve to get $p=2, q=\frac{3}{2}$ (discard);
When $m=3, n=1$, from equations (1) and (2), we solve to get $p=5, q=3$.
In summary, $p+q=8$.