Example 14 Find the number of pairwise non-congruent triangles with vertices at the vertices of a regular 2000-gon.

保留源文本的换行和格式，翻译结果如下：

Example 14 Find the number of pairwise non-congruent triangles with vertices at the vertices of a regular 2000-gon.
Solution: The vertices of a regular 2000-gon divide its circumcircle into 2000 equal parts. Let the arcs between the three vertices of a triangle contain $a, b, c$ parts, respectively, then $a+b+c=2000$. Now we classify the triangles based on their shapes:
(1) The triangle is equilateral. Since 2000 is not a multiple of 3, such a triangle does not exist.
(2) The triangle is isosceles. Without loss of generality, let $a=b$, then $c=2000-2a$, and $a$ can take values $1,2, \cdots, 999$. There are 999 such triangles.
(3) The triangle has three unequal sides. Let $1 \leqslant a < 1000, a+2b < a+b+c = 2000$, i.e., $1 \leqslant a < b, 1000-b < a < 2000-2b$.

Conversely, for integers $a, b$ satisfying condition (1), let $c=2000-a-b$, then the triangle with vertices separated by arcs of $a, b, c$ parts is a triangle with three unequal sides.
When $b=i (501 \leqslant i \leqslant 666)$, there are $2i-1001$ such $a$.
When $b=i (667 \leqslant i \leqslant 998)$, there are $999-i$ such $a$.
For other $b$, the corresponding $a$ does not exist. Thus, the total number of such triangles is
$$
\sum_{i=501}^{666}(2i-1001) + \sum_{i=667}^{998}(999-i) = 82834.
$$

When the triangle is not acute, $c \geqslant 1000$, hence $a+b=2000-c \leqslant 1000$. When $a+b=2k (k=1,2, \cdots, 500)$, there are $k-1$ pairs of $a, b$ such that $1 \leqslant a < b$. When $a+b=2k+1 (k=1,2, \cdots, 499)$, there are $k$ pairs of $a, b$ such that $1 \leqslant a < b$. Therefore, the total number of such triangles is $2(1+2+\cdots+499)=249500$.
In summary, the total number of triangles is
$$
999 + 82834 + 249500 = 333333.
$$