4. If $p, q \in N^{+}$ and $p+q>2017, 0<p<q \leq 2017, (p, q)=1$, then the sum of all fractions of the form $\frac{1}{p q}$ is $\qquad$
4. $\frac{1}{2}$

Analysis: Generalize 2017 to a more general case: If $p, q \in N^{+}$ and $p+q>n, 0<p<q \leq n, (p, q)=1$, then the sum of all fractions of the form $\frac{1}{p q}$ is $\frac{1}{2}$; We denote the sum of all such fractions as $S_{n}$. When $n=2$, it is clearly true. When $n=k+1$, let the set of all fractions of the form $\frac{1}{p q}$ (where $p+q=k, (p, q)=1$) be denoted as $A$, and the sum of these fractions as $S$. Let the set of all fractions of the form $\frac{1}{k p}$ (where $p<k, (p, k)=1$) be denoted as $B$, and the sum of these fractions as $T$. Then, $S_{k+1}=S_{k}+T-S$;

Next, we prove: $T=S$.
For any fraction $\frac{1}{p q}$ in $A$, there is exactly one pair of fractions $\frac{1}{p k}, \frac{1}{q k}$ in $B$ corresponding to it, and $\frac{1}{p k}+\frac{1}{q k}=\frac{1}{p q}$. This correspondence is one-to-one, hence $T=S$, i.e., $S_{2017}=S_{2}=\frac{1}{2}$.