2. Find the smallest natural number $l$, such that in a graph with $n(n \geqslant 3)$ vertices, if the number of edges is more than $\frac{1}{2} n \sqrt{n-1}$, then there is a cycle of length less than $l$.
2. Solution: (1) There must be a triangle or a 4-cycle in the graph.

Assume not, then consider the number of triples $\left(P_{i}, P_{j}, P_{k}\right)$ where $P_{i} P_{j}, P_{i} P_{k} \in E, i, j, k$ are distinct, and $1 \leqslant i, j, k \leqslant n$.

For a fixed $\left(P_{i}, P_{k}\right)$, there is at most one $P_{j}$ that forms a triple with it, such that when $P_{i} P_{k} \in E$, such a triple does not exist.

For a fixed $P_{i}$, there are $\mathrm{C}_{d_{i}}^{2}$ triples, and $\sum_{i=1}^{n} d_{i}> n \sqrt{n-1}$, thus
$\mathrm{C}_{n}^{2}-\frac{1}{2} n \sqrt{n-1}>\sum_{i=1}^{n} \mathrm{C}_{d_{i}}^{2} \geqslant n \cdot \frac{\sqrt{n-1}(\sqrt{n-1}-1)}{2}$, which leads to
$\frac{n-1}{2}-\frac{\sqrt{n-1}}{2}>\frac{n-1-\sqrt{n-1}}{2}$, a contradiction.
Therefore, the graph must have a 3-cycle or a 4-cycle.
(2) When $n \geqslant 4$, construct a bipartite graph $G(X, Y), |X| = \left[\frac{n}{2}\right], |Y| = \left[\frac{n+1}{2}\right]$, with any two points between $X$ and $Y$ connected, at this point $|E| = \left[\frac{n^{2}}{4}\right] > \frac{1}{2} n \sqrt{n-1}$, and the graph has no triangles.
(3) When $n=3$, $\frac{1}{2} n \sqrt{n-1} > 2$, so there is a triangle. $\square$
In summary, $l_{\min }=\left\{\begin{array}{l}4, n=3 ; \\ 5, n \geqslant 4 .\end{array}\right.$