Six, there are 6000 points inside a circle, and no three points are collinear:
(1) Can this circle be divided into 2000 parts, each containing exactly three points? How to divide it?
(2) If the three points in each part satisfy: the distance between any two points is an integer and does not exceed 9, then using the three points in each part as vertices to form triangles, how many of these triangles are of the same size at least?
(1) 6000 points inside a circle can determine $C_{6000}^{2}$ lines. Since $C_{6000}^{2}$ is a finite number, there must exist a tangent line to the circle that is not parallel to any of the $C_{5000}^{2}$ lines, denoted as $l$. Moving $l$ parallel within the circle, it is clear that the 6000 points will be crossed (if it crosses two points simultaneously, then the line connecting these two points must be parallel to $l$, which contradicts the choice of $l$). Specifically, when $l$ crosses 3 points and has not yet crossed the fourth point, a chord $l_{1}$ is drawn.

Similarly, when crossing the 4th, 5th, 6th points, chords $l_{2}, l_{3}, l_{4}, \cdots$ are drawn. In this way, 1999 chords can be drawn, dividing the circle into 2000 segments, each containing three points. Each set of three points forms a triangle, resulting in 2000 triangles.
(2) It can be calculated that the number of triangles with integer side lengths and the longest side not exceeding 9 is 95. Let the side lengths be $a \leqslant b \leqslant c \leqslant 9$, where $a, b, c$ are all integers.
When $c=9$, and $a+b>c$, we have
\begin{tabular}{|c|c|c|}
\hline Values of $b$ & Values $a$ can take & \begin{tabular}{l} 
Number of triangles
\end{tabular} \\
\hline 5 & 5 & 1 \\
\hline 6 & $4,5,6$ & 3 \\
\hline 7 & $3,4,5,6,7$ & 5 \\
\hline 3 & $\mid 2,3,4,5, \cdots, 8$ & 7 \\
\hline 9 & $2,3,4,5, \cdots, 9$ & 9 \\
\hline
\end{tabular}

Thus, when $c=9$, 25 different triangles can be obtained. When $c=8$, and $a+b>c$, we have

Thus, when $c=8$, 20 different triangles can be obtained. Similarly, when $c=7,6,5,4,3$, 2,1, the number of different triangles are 16, 12, 9, 6, 4, 2, 1, respectively.

Therefore, the total number of different triangles with integer side lengths and the longest side not exceeding 9 is
$$
1+2+4+6+9+12+16+20+25=95
$$

Thus, among the 2000 triangles, the number of triangles that are exactly the same in size is at least
$$
\left[\frac{2000}{95}\right]+1=22 \text { triangles. }
$$
(Supplied by Chen Dingchang, Zhejiang)