Example 4 Let integer $n \geqslant 3, a_{1}, a_{2}, \cdots, a_{n}$ be any $n$ distinct real numbers with a positive sum; $b_{1}, b_{2}, \cdots, b_{n}$ is a permutation of these $n$ numbers. If for $k=1,2, \cdots, n$, we have $\sum_{i=1}^{k} b_{i}>0$, then $b_{1}, b_{2}, \cdots, b_{n}$ is called a "good permutation". Find the minimum number of good permutations.

[Analysis] According to the definition of a good permutation, for $k=$ $1,2, \cdots, n$, we need $\sum_{i=1}^{k} b_{i}>0$.

Therefore, the fewer positive numbers in $a_{1}, a_{2}, \cdots, a_{n}$, the fewer choices there are for the first few terms in the sequence $b_{1}, b_{2}, \cdots, b_{n}$, and thus the total number of good permutations might be smaller. The most extreme case is when there is only one positive number in $a_{1}, a_{2}, \cdots, a_{n}$.
Let $a_{1}>0, a_{2}, a_{3}, \cdots, a_{n}$ all be less than 0. In this case, it is easy to see that a good permutation must have $a_{1}$ in the first position, and the order of the last $n-1$ positions $a_{2}, a_{3}, \cdots, a_{n}$ is not required. Therefore, the number of good permutations is $(n-1)!$.
To prove: $(n-1)!$ is the minimum number of good permutations.
For convenience, we generalize the definition of a good permutation in the problem. If for $i=1,2, \cdots, k$, we have $\sum_{j=1}^{i} a_{j}>0$, then the permutation $a_{1}, a_{2}, \cdots, a_{k}(k \leqslant n)$ is called a "k-element good permutation".

It is easy to see that the good permutation required in the problem is an n-element good permutation. The good permutations mentioned below include any k-element good permutations.

For any $a_{1}, a_{2}, \cdots, a_{n}$ that satisfy the conditions, place $a_{1}$, $a_{2}, \cdots, a_{n}$ on a circle, and the number of circular permutations is $(n-1)!$. We only need to prove: For any circular permutation, there is at least one corresponding n-element good permutation, and different circular permutations correspond to different n-element good permutations.

Let $a_{1}, a_{2}, \cdots, a_{n}$ have a circular permutation $c_{1}, c_{2}$, $\cdots, c_{n}$ (with the convention $c_{n+i}=c_{i}$). Consider the set $A$ of all good permutations generated by the circular permutation $c_{1}, c_{2}$, $\cdots, c_{n}$ (a positive number is a one-element good permutation, so this set is non-empty). Since $A$ is a finite set, according to the extreme principle, take the longest good permutation in it, and assume that this good permutation is an $l$-element good permutation $c_{1}$, $c_{2}, \cdots, c_{l}$. We want to prove $l=n$.
(1) If $l=n$, then the conclusion is proved.
(2) If $l < n$, then
$$
\begin{array}{l}
\sum_{j=1}^{l} c_{j}>0 \\
\Rightarrow c_{l+1}+c_{l+2}+\cdots+c_{n}<0 \\
\Rightarrow c_{l+2}+c_{l+3}+\cdots+c_{n}>0 .
\end{array}
$$

Let $t$ be the smallest $t$ such that $c_{t}+c_{t+1}+\cdots+c_{n}>0$. Then $c_{t}, c_{t}+c_{t+1}, \cdots, c_{t}+c_{t+1}+\cdots+c_{n-1}$ are all positive. Hence, $c_{t}, c_{t+1}, \cdots, c_{n}$ is an $(n-t+1)$-element good permutation. Therefore, $c_{t}, c_{t+1}, \cdots, c_{n}, c_{1}, c_{2}, \cdots, c_{l}$ is a good permutation longer than $l$, which is a contradiction (if $c_{t}, c_{t+1}, \cdots, c_{n}$ overlaps with $c_{1}, c_{2}, \cdots, c_{l}$, then removing the overlapping items from $c_{1}, c_{2}, \cdots, c_{l}$ still results in a good permutation longer than $l$). Therefore, $l=n$.

In summary, a circular permutation corresponds to at least one good permutation, and different circular permutations correspond to different good permutations. Therefore, the number of good permutations is at least $(n-1)!$.