Three. (50 points) There are 2014 points distributed on a circle, which are arbitrarily colored red and yellow. If starting from a certain point and moving around the circle in any direction to any position, the number of red points (including the starting point) is always greater than the number of yellow 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 yellow points on the circle.

保留源文本的换行和格式，直接输出翻译结果如下：

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Three. (50 points) There are 2014 points distributed on a circle, which are arbitrarily colored red and yellow. If starting from a certain point and moving around the circle in any direction to any position, the number of red points (including the starting point) is always greater than the number of yellow 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 yellow points on the circle.
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Three, the advantages must be red points.
First, consider the simple case.
When there are $1, 2, 3, 4, 5, 6, 7$ points on the circumference, if there is at least one advantage point on the circumference, then the maximum number of yellow points on the circumference are $0, 0, 0, 1, 1, 1, 2$. From this, we can derive a general conclusion:

When there are $3n+1$ points on the circumference, and they are arbitrarily colored red and yellow. To ensure that there is at least one advantage point on the circumference, the number of yellow points must not exceed $n$.
Next, we will prove this using mathematical induction.
When $n=1$, there are four points on the circumference (one yellow point and three red points). Among the three consecutive red points, take the middle one, and it is easy to see that it is an advantage point. Therefore, the proposition holds for $n=1$.

Assume that the proposition holds for $n=k$, i.e., when there are $3k+1$ points on the circumference, colored red and yellow, to ensure that there is at least one advantage point on the circumference, the number of yellow points must not exceed $k$.

When $n=k+1$, take any one of the $k+1$ yellow points, denoted as $A$, and take the nearest red points on either side of $A$, denoted as $B$ and $C$. Remove these three points from the circumference, leaving $3k+1$ points, which satisfy the conditions of the proposition for $n=k$.

By the induction hypothesis, there is at least one advantage point on the circumference, denoted as $P$. Now, put $A, B, C$ back to their original positions.
We need to prove that point $P$ is still an advantage point.
Since $P$ is a red point, $P$ must be outside the arc $\overparen{BAC}$. Therefore, from point $P$ to any position on the arc $\overparen{BA}(\overparen{BC})$, the difference between the number of red points and yellow points is at least 1, and it is at least 0 when reaching point $A$. Hence, $P$ is still an advantage point.
Thus, the proposition holds for $n=k+1$.
On the other hand, if the number of yellow points among the $3n+1$ points is $n+1$, divide the circumference into $n+1$ segments, and place $2n$ red points into each segment, with at most two points in each segment, then no red point can be an advantage point.
Therefore, the maximum number of yellow points on the circumference is $n$.
Since $2014 = 3 \times 671 + 1$, the maximum number of yellow points on the circumference is 671.