I+ Place nine balls numbered $1,2, \cdots, 9$ randomly on nine equally spaced points on a circle, with one ball on each point. Let the sum of the absolute differences of the numbers of all adjacent balls on the circle be $S$. Find the probability of the arrangement that minimizes $S$. (Note: If one arrangement can be made to coincide with another by rotation or reflection, they are considered the same arrangement)
To place nine differently numbered small balls on the nine equally divided points of a circle, with one ball on each point, is equivalent to a circular permutation of nine different elements. Therefore, there are $8!$ ways to place them. Considering the factor of flipping, the number of essentially different ways is $\frac{8!}{2}=4 \cdot 7!$.

Next, we seek the number of ways to achieve the minimum value of $S$: On the circle, from 1 to 9, there are two paths, the major arc and the minor arc. For any one of these paths, let $x_{1}, x_{2}, \cdots, x_{k}$ be the ball numbers sequentially arranged on this arc segment. Then,
$$
\begin{aligned}
& \left|1-x_{1}\right|+\left|x_{1}-x_{2}\right|+\cdots+\left|x_{k}-9\right| \\
\geqslant & \left|\left(1-x_{1}\right)+\left(x_{1}-x_{2}\right)+\cdots+\left(x_{k}-9\right)\right| \\
= & |1-9|=8 .
\end{aligned}
$$

The above inequality holds with equality if and only if $1<x_{1}<x_{2}<\cdots<x_{k}<9$, i.e., the ball numbers on each arc segment are arranged in increasing order from 1 to 9.
Therefore,
$$
S_{\text {minimum }}=2 \cdot 8=16.
$$

From the above, we know that once the ball numbers $\left\{1, x_{1}, x_{2}, \cdots x_{k}, 9\right\}$ on each arc segment are determined, the sorting scheme that achieves the minimum value is uniquely determined.
In the set $\{1,2, \cdots, 9\}$, excluding 1 and 9, there are 7 remaining ball numbers $\{2,3, \cdots, 8\}$. Dividing them into two subsets, the subset with fewer elements has $\mathrm{C}_{7}^{0}+\mathrm{C}_{7}^{1}+\mathrm{C}_{7}^{2}+\mathrm{C}_{7}^{3}=2^{6}$ cases. Each case corresponds to a unique arrangement on the circle that achieves the minimum $S$ value, i.e., the number of favorable events is $2^{6}$. Therefore, the required probability is
$$
P=\frac{2^{6}}{4 \cdot 7!}=\frac{1}{315}.
$$
20 points