1. Let $p, q \in N$, and $1 \leqslant p<q \leqslant n$, where $n$ is a natural number not less than 3. Then the sum of all fractions of the form $\frac{p}{q}$ is $\qquad$ .
$=1 \cdot \frac{1}{4} n(n-1)$.
Group all fractions of the form $\frac{p}{q}$ as follows:
$$
\left(\frac{1}{2}\right),\left(\frac{1}{3}, \frac{2}{3}\right), \cdots,\left(\frac{1}{n}, \frac{2}{n}, \cdots, \frac{n-1}{n}\right) \text {, }
$$

where the $k$-th group contains $k$ fractions. It is not difficult to find that the sum of the $k$ elements in the $k$-th group $(1 \leqslant k \leqslant n-1)$ is
$$
\frac{1}{k+1}(1+2+\cdots+k)=\frac{k}{2} .
$$

Therefore, the sum of all fractions of the form $\frac{p}{q}$ is
$$
s=\frac{1}{2}+\frac{2}{2}+\frac{2}{2}+\cdots+\frac{n-1}{2}=\frac{1}{4} n(n-1) .
$$