Let the non-empty subset family $\left\{A_{1}, A_{2}, A_{3}\right\}$ be a partition of the set $M=\{1,2, \cdots, 10\}$, i.e., the non-empty sets $A_{1}$, $A_{2}$, $A_{3}$ are pairwise disjoint, and their union is $M$. If there exists a permutation $i_{1}, i_{2}, i_{3}$ of $1,2,3$ such that
$$
\max A_{i_{k}}>\min A_{i_{k+1}}, \quad k=1,2,3,
$$

where it is defined that $i_{4}=i_{1}$, then $\left\{A_{1}, A_{2}, A_{3}\right\}$ is called a good partition of $M$. Find the number of all good partitions of $M$.
Let's assume $\min A_{1}\min A_{2}$.
(1) When $\max A_{2}>\max A_{3}$, take $i_{1}=1, i_{2}=2, i_{3}=3$, then $\max A_{i_{k}}>\min A_{i_{k+1}}, k=1,2,3$ $\left(i_{4}=i_{1}\right)$.
(2) When $\max A_{2}\max A_{1}$ and $\min A_{3}>\min A_{2}$, then it is clear that such a partition is not good. Therefore, $\min A_{3}\min A_{i_{k+1}}, k=1,2,3\left(i_{4}=i_{1}\right)$; if $\min A_{3}\min A_{i_{k+1}}, k=1,2,3\left(i_{4}=i_{1}\right)$.

In summary, the non-empty family of sets $A_{1}, A_{2}, A_{3}$ is a good partition of set $M$ if and only if: $A_{1}$, $A_{2}, A_{3}$ are not sets of the form $\{1,2, \cdots, n\}(n=1,2, \cdots, 8)$ and $\{m, \cdots, 10\}(m=3,4, \cdots, 10)$.
Let
$$
\begin{aligned}
P & =\left\{\left\{A_{1}, A_{2}, A_{3}\right\} \mid \text { non-empty sets } A_{1}, A_{2}, A_{3} \text { are a partition of } M \right\}, \\
P_{1} & =\left\{\left\{A_{1}, A_{2}, A_{3}\right\} \mid\left\{A_{1}, A_{2}, A_{3}\right\} \in P, A_{1}=\{1,2, \cdots, n\}(n=1,2, \cdots, 8)\right\}, \\
P_{2} & =\left\{\left\{A_{1}, A_{2}, A_{3}\right\} \mid\left\{A_{1}, A_{2}, A_{3}\right\} \in P, A_{3}=\{m, \cdots, 10\}(m=3,4, \cdots, 10)\right\} .
\end{aligned}
$$

To calculate $|P|$, first calculate the number of non-empty ordered solutions to the system of set equations
$$
\left\{\begin{array}{l}
A_{1} \cup A_{2} \cup A_{3}=M, \\
A_{i} \cap A_{j}=\varnothing \quad(1 \leqslant i \neq j \leqslant 3)
\end{array}\right.
$$

Since each number $s(s=1,2, \cdots, 10)$ in $M$ belongs exactly to one of $A_{1}, A_{2}, A_{3}$, the number of ordered solutions to the system (1) is $3^{10}$, but this includes cases where one set is empty and where two sets are empty. There are $3\left(2^{10}-2\right)$ cases where exactly one set is empty, and 3 cases where exactly two sets are empty. Therefore, the number of non-empty ordered solutions to the system (1) is $3^{10}-3\left(2^{10}-2\right)-3$. Thus, $|P|=\frac{1}{6}\left(3^{10}-3\left(2^{10}-2\right)-3\right)$.
It is also easy to see that
$$
\begin{array}{l}
\left|P_{1}\right|=\left|P_{2}\right|=\frac{1}{2}\left(2^{9}-2\right)+\frac{1}{2}\left(2^{8}-2\right)+\cdots+\frac{1}{2}\left(2^{2}-2\right)=2^{9}-10, \\
\left|P_{1} \cap P_{2}\right|=1+2+\cdots+8=36 .
\end{array}
$$

By the principle of inclusion-exclusion, the number of good partitions is
$$
\begin{aligned}
|P|-\left(\left|P_{1}\right|+\left|P_{2}\right|\right)+\left|P_{1} \cap P_{2}\right| & =\frac{1}{6}\left(3^{10}-3\left(2^{10}-2\right)-3\right)-2\left(2^{9}-10\right)+36 \\
& =\frac{3^{9}+1}{2}-3 \cdot 2^{9}+56=9842-1536+56=8362 .
\end{aligned}
$$