3. The answer is $k=1512$. This number is achieved in the example for vertices $A$ and $B$ in the figure.

![](https://cdn.mathpix.com/cropped/2024_06_04_fc1ffcc45720460ecdffg-2.jpg?height=200&width=997&top_left_y=2341&top_left_x=541)

Let's prove the statement for $k=1512$. Let $s=A_{0} A_{1} A_{2} \ldots A_{n}$ be the shortest path from vertex $A=A_{0}$ to $B=A_{n}$. For $1 \leqslant i \leqslant n-1$, vertex $A_{i}$ is connected to at least one more vertex $B_{i}$, in addition to $A_{i-1}$ and $A_{i+1}$. Vertices $A_{0}$ and $A_{n}$ are connected to at least two more vertices: $B_{0}$ and $C_{0}$, and $B_{n}$ and $C_{n}$, respectively.

None of the vertices $B_{i}, C_{0}$, or $C_{n}$ coincide with $A_{j}$ for $j \neq i-1, i+1$, otherwise the path $s$ could be shortened. Similarly, all $\left[\frac{n-3}{3}\right]+4$ vertices $B_{0}, C_{0}, B_{3}, B_{6}, \ldots, B_{3\left[\frac{n-3}{3}\right]}, B_{n}, C_{n}$ are distinct, otherwise the path $s$ could be shortened. In total, the graph $G$ has at least $n+1+\left[\frac{n-3}{3}\right]+4 \geqslant \frac{4 n+10}{3}$ vertices. It follows that $\frac{4 n-10}{3} \leqslant 2020$, i.e., $n \leqslant 1512$.
3. Find the smallest natural number $k$ for which the following statement holds:

If $G$ is a simple connected graph with 2020 vertices in which every vertex has a degree of at least 3, then from any vertex to any other vertex, one can reach by a path of length at most $k$.

