7. Let the set of $n$ points $V=\left\{A_{0}, A_{1}, A_{2}, \cdots, A_{n-1}\right\}$ be the universal set, and let the set of all neighbors of $A_{i}$ (points connected to $A_{i}$ by a line segment) be denoted as $B_{i}$. The number of points in $B_{i}$ is denoted as $\left|B_{i}\right|=b_{i}$.
Clearly, $\sum_{i=0}^{n-1} b_{i}=2 l$ and $b_{i} \leqslant n-1(i=0,1,2, \cdots, n-1)$.
If there exists $b_{i}=n-1(i=0,1,2, \cdots, n-1)$, then we only need to take
$$\iota=(n-1)+\left[\frac{n-1}{2}\right]+1 \leqslant \frac{1}{2}(q+1)(n-1)+1=\frac{1}{2} q(q+1)^{2}+1 .$$

Then the graph must contain a quadrilateral.
Therefore, we only need to discuss $b_{i}0$.
From (2), (3), and $\left(n-b_{0}\right)(q+1),\left(n-b_{0}-1\right) q$ being positive integers, we get
$$\left(n q-q+2-b_{0}\right)\left(n q-q-n+3-b_{0}\right)>q(q+1)\left(n-b_{0}\right)\left(n-b_{0}-1\right) .$$

This contradicts equation (1), hence the original proposition holds.