Three. (50 points)
In a square with a side length of 4, 2002 points are placed. How many pairs of points are there with a distance less than 3 (find the largest specific number)?
First, among any 5 points, there must be a pair of points with a distance less than 3. This is because: dividing the square into 4 smaller $2 \times 2$ squares, by the pigeonhole principle, at least 2 points must fall into the same smaller square, and the distance between two points in the same smaller square is no more than $2 \sqrt{2}n-4 \text{ ) }
$$
3. If the complement graph $G_{n}$ has no $K_{3}$, and has $K_{2}$, then by the same reasoning as 1 and 2, it can be proven. In this case, the number of edges is no less than the number of edges in 1.
4. If the complement graph $G_{n}$ has no $K_{2}$, then $G_{n}$ is $K_{n}$, and the number of edges is $C_{n}^{2}>\sum_{i=1}^{4} C_{n u}^{2}$.

Combining 1, 2, 3, and 4, the induction holds.
Therefore, the number of edges in $G_{2002}$ is no less than
$$
2 C_{500}^{2}+2 C_{501}^{2}=500000
$$

That is, among 2002 points, there are at least 500000 pairs of points with a distance less than 3.
Furthermore, taking a small square with side length $\frac{1}{2}$ at each of the 4 corners of the square, and placing 500, 500, 501, and 501 points in them respectively, the distance between points in different small squares is no less than 3. Therefore, the number of pairs of points with a distance less than 3 is only $2 C_{500}^{2}+2 C_{501}^{2}=500000$ pairs.
In summary, the number of pairs of points with a distance less than 3 is at least 500000 pairs.