Let $O$ be the center of circle $C$, $a_n$ is the length of a regular $n$-gon $A_1A_2\cdots A_n$ inscribed in $\odot O$. Then $a_6 = 10 > 9$, $a_7 < 10 \times \frac{2\pi}{7} < 10 \times \frac{2 \times 3.15}{7} < 9$.

(1) Let $A_1A_2\cdots A_6$ be a regular 6-gon inscribed in $\odot O$, then $A_iA_{i+1} = a_6 > 9$. So we can choose a point $B_i$ in $\widehat{A_iA_{i+1}}$ such that $B_iA_{i+1} > 9$. Then $\angle B_iOA_{i+1} > \frac{2\pi}{7}$ ($A_7 = A_1$), and
$$\angle A_iOB_i = \angle A_iOA_{i+1} - \angle B_iOA_{i+1} < \frac{2\pi}{6} - \frac{2\pi}{7} < \frac{2\pi}{7}.$$
It follows that $A_iB_i < 9$ ($i = 1, 2, \dots, 6$).

In each of $\widehat{A_1B_1}, \widehat{A_2B_2}, \widehat{A_3B_3}$, choose 11 points, and in each of $\widehat{A_4B_4}, \widehat{A_5B_5}, \widehat{A_6B_6}$, choose 10 points. We obtain a set $M$ which has 63 points on the circle $C$. It is easy to see for $M$ the value of $S$ is $S_0$, where
$$\begin{aligned} S_0 &= \binom{3}{3} \times 11^3 + \binom{3}{2} \cdot \binom{3}{1} \times 11^2 \times 10 \\ &\quad + \binom{3}{1} \cdot \binom{3}{2} \times 11 \times 10^2 + \binom{3}{3} \times 10^3 \\ &= 23\,121. \end{aligned}$$
So the maximum value of $S$ is not less than $S_0$.

(2) We prove that the maximum is $S_0$. We need three lemmas.

Lemma 1 For $P$ on circle $C$, we call the arc $\widehat{APB}$ “an arc of $P$”, if $P$ is the midpoint of arc $\widehat{APB}$, and $\angle AOB = \frac{4\pi}{7}$. Now for every given $n$ points on circle $C$, there is a point $P$, such that there are $\lfloor \frac{n+5}{6} \rfloor$ points of the given $n$ points on the “arc of $P$”.

Proof: Let $A$ be one of the given $n$ points, and an “arc of $A$” be $\widehat{A_1AA_6}$. Now suppose $A_2, A_3, \dots, A_5$ are on the arc $\widehat{A_1A_6}$ (not including $A$), and $A_1A_2 = A_2A_3 = \dots = A_5A_6$ (see the figure). So $\angle A_iOA_{i+1} = \frac{2\pi}{7}, i = 1, 2, \dots, 5$.

If there is a point $P_i$ (of the given $n$ points) on $\widehat{A_iA_{i+1}}$, then all the points (of the given $n$ points) on $\widehat{A_iA_{i+1}}$ are on “an arc of $P_i$”. So the given $n$ points are on $6$ “arcs of $P_i$” (including “arc of $A$”). This shows that there are $\lfloor \frac{n-1}{6} \rfloor + 1 = \lfloor \frac{n+5}{6} \rfloor$ points of the given points on an “arc of $P_i$”, where $P_i$ is one of the given $n$ points.

Lemma 2 Take the arc $\widehat{A_1BA_6}$ arbitrary on the circle $C$ with radius 10, where $\widehat{A_1BA_6}$ is $\frac{5}{7}$ of the perimeter. Then take any $5m+r$ points on the arc $\widehat{A_1BA_6}$ ($m, r$ are non-negative integers and $0 \le r < 5$). Prove that number of lines from the given points whose lengths are more than 9 is at most
$$10m^2 + 4rm + \frac{1}{2}r(r-1).$$

Proof: Divide $\overrightarrow{A_1BA_6}$ in five equal parts, where the corresponding points are $A_2, A_3, A_4, A_5$ (see the figure), then the length of $\overrightarrow{A_iA_{i+1}}$ is exactly $\frac{1}{7}$ of the perimeter ($i = 1, 2, 3, 4, 5$), and the distance of any two points is not more than $a_7 < 9$. Suppose there are $m_i$ given points on the arc $\overrightarrow{A_iA_{i+1}}$, then the number of lines from the given points whose lengths are more than 9 is at most
$$l = \sum_{1 \le i < j \le 5} m_i m_j, \qquad \textcircled{1}$$
where
$$m_1 + m_2 + \cdots + m_5 = 5m + r. \qquad \textcircled{2}$$
Since there are finitely many non-negative integer groups ($m_1, m_2, m_3, m_4, m_5$), the maximum value of $l$ exists. Now we prove that when the maximum is attained the inequality
$$|m_i - m_j| \le 1, (1 \le i < j \le 5),$$
must hold.

In fact, if there exist $i, j$ ($1 \le i < j \le 5$) such that $|m_i - m_j| \ge 2$ when the maximum is attained, we can suppose $m_1 - m_2 \ge 2$. Then let
$$m_1' = m_1 - 1,\quad m_2' = m_2 + 1,\quad m' = m,$$
and the corresponding integer is $l'$, we will have
$$\begin{align*} m_1' + m_2' &= m_1 + m_2, \\ m_1' + m_2' + m_3' + m_4' + m_5' &= m_1 + m_2 + m_3 + m_4 + m_5, \\ l' - l &= (m_1'm_2' - m_1m_2) + [(m_1' + m_2') \\ &\quad -(m_1 + m_2)](m_3 + m_4 + m_5) \\ &= m_1 - m_2 - 1 \ge 1. \end{align*}$$
Contradiction!

Therefore, when $l$ reaches the maximum value, the number of $m+1$ is $r$ and the number of $m$ is $5-r$. Thus, the number of lines from the given points whose lengths are more than 9 are at most
$$\begin{aligned} & \binom{r}{2}(m+1)^2 + \binom{r}{1}(5-r)(m+1)m + \binom{5-r}{2}m^2 \\ &= 10m^2 + 4rm + \frac{1}{2}r(r-1). \end{aligned}$$

Lemma 3 Take arbitrary $n$ points on the circle $C$ with radius 10 to form set $M$, where $n = 6m + r$ ($m, r$ are non-negative integers, $0 \le r < 6$). Assume that there are $S_n$ triangles whose vertices are from $M$ and each side is longer than 9. Prove
$$S_n \le 20m^3 + 10rm^2 + 2r(r-1)m + \frac{1}{6}r(r-1)(r-2).$$

Proof: We shall prove by mathematical induction.

When $n = 1, 2$, $S_n = 0$. It is true.

Suppose when $n = k$, it is true and set $k = 6m + r$ ($m, r$ are non-negative integers, $0 \le r < 6$). Then
$$S_k \le 20m^3 + 10rm^2 + 2r(r-1)m + \frac{1}{6}r(r-1)(r-2).$$
From Lemma 1, when $n = k + 1$, the $k + 1$ given points must include the point $P$, where at least $\lfloor \frac{k+1+5}{6} \rfloor = m + 1$ given points are in the $\frac{2}{7}\text{arc } A_1PA_6$. And the distances of such points to $P$ are $\le PA_1 = PA_6 = a_7 < 9$. Hence, there are at most $(k+1)-(m+1) = 5m+r$ given points whose distances to $P$ are more than 9, and such points are all in the other $\frac{5}{7}\arccos \overline{A_1PA_6}$ without $P$. From Lemma 2, the lines from such points whose lengths are more than 9 are at most
$$10m^2 + 4rm + \frac{1}{2}r(r-1).$$
(From Lemma 2, when $r=5$, it is $10(m+1)^2$, which is also true.) Thus, the number of triangles whose vertex is $P$ and each side is larger than 9 is not more than
$$S_p = 10m^2 + 4rm + \frac{1}{2}r(r-1).$$
Without $P$, there are $k = 6m + r$ given points. Let there be $S_k$ triangles whose vertices are from the $k$ points and each side is larger than 9, then using mathematical induction, we get
$$S_k \le 20m^3 + 10rm^2 + 2r(r-1)m + \frac{1}{6}r(r-1)(r-2).$$
Furthermore
$$\begin{align*} S_{k+1} &= S_k + S_p \\ &\le 20m^3 + 10rm^2 + 2r(r-1)m + \frac{1}{6}r(r-1)(r-2) \\ &\quad + 10m^2 + 4rm + \frac{1}{2}r(r-1) \\ &= 20m^3 + 10(r+1)m^2 + 2r(r+1)m \\ &\quad + \frac{1}{6}r(r-1)(r+1), \end{align*}$$
which means the case $n = k + 1 = 6m + (r + 1)$ is also true.

On the other hand, when $r=5$, then $m = k + 1 = 6(m+1)$ and $S_{k+1}$ can be simplified to $S_{k+1} = 20(m+1)^3$, which is also true.

Therefore, we have proved Lemma 3.

Now considering the original problem, we have
$$n = 63 = 6 \times 10 + 3.$$
It follows from Lemma 3,
$$\begin{aligned} S &\le 20 \times 10^3 + 10 \times 10^2 + 2 \times 3 \times 2 \times 10 + \frac{1}{6} \times 3 \times 2 \times 1 \\ &= 23\,121. \end{aligned}$$
Thus, $S_{\max} = 23\,121$.