Three. (50 points) There are 2002 points distributed on a circle. Now, they are arbitrarily colored white or black. If starting from a certain point and moving in any direction around the circle to any point, the total number of white points (including the point itself) is always greater than the number of black points, then the point is called a good point. To ensure that there is at least one good point on the circle, find the maximum number of black points that can be colored.
From the problem, we know that a good point must be white. The following discussion is for the general case: there are $3n+1$ points on the circumference, which are colored black and white. Only when the number of black points $\leqslant n$, can we ensure that there must be a good point.
We prove this by mathematical induction on $n$.
(1) When $n=1$, there are 4 points on the circumference. The black and white coloring is one black point and three white points. Among the three consecutive white points, take the middle one, and it is easy to see that it is a good point. Therefore, the conclusion holds when $n=1$.
(2) Suppose the proposition holds when $n=k$. Then, when $n=k+1$, take any one of the $k+1$ black points and denote it as $A$. Take the nearest white points to $A$ on both sides and denote them as $B$ and $C$. Remove these three points $A$, $B$, and $C$ from the circumference, leaving $3k+1$ points, of which $k$ are black. By the induction hypothesis, there must be a good point among these $3k+1$ points, denoted as $D$ (a white point). Then, put $A$, $B$, and $C$ back onto the circumference, resulting in $3(k+1)+1$ points.
Now we prove that $D$ is still a good point.
Indeed, since $D$ is a white point, point $D$ must be outside the arc $\overparen{BAC}$. Therefore, when moving from point $D$ along the points on the circumference to reach the points within $\overparen{BA}$ (or $\overparen{CA}$) (excluding point $A$), the difference between the number of white points and black points is increased by 1 compared to the original difference. Thus, when reaching point $A$, the difference between the number of white points and black points must be greater than 0, which means $D$ is still a good point. Therefore, the conclusion holds when $n=k+1$.

On the other hand, when the total number of black points is $n+1$, there is a black and white coloring such that no good point exists: $n+1$ black points divide the circumference into $n+1$ small arcs. Then, place the remaining $2n$ white points into these $n+1$ small arcs so that no arc has more than 2 white points, which is always possible. This coloring ensures that no good point exists.
$$
\because 2002=3 \times 667+1 \text {, }
$$
$\therefore$ To ensure the existence of a good point, the total number of black points $\leqslant 667$. In summary, the maximum number of black points is 667.
(Xu Jingling, Anhui Province, Susong Middle School, 246501)