Example 5 Let $n \geqslant 3$ be a positive integer, and $a_{1}, a_{2}, \cdots, a_{n}$ be any $n$ distinct real numbers with a positive sum. If a permutation $b_{1}, b_{2}, \cdots, b_{n}$ of these numbers satisfies: for any $k=1,2, \cdots, n$, $b_{1}+b_{2}+\cdots+b_{k}>0$, then this permutation is called good. Find the minimum number of good permutations. (2002


Analysis and Solution: Consider the worst-case scenario: for $k=1,2, \cdots, n, b_{1}+b_{2}+\cdots+b_{k}>0$ is difficult to satisfy. This only requires that the number of negative numbers in $a_{1}, a_{2}, \cdots, a_{n}$ is as many as possible. Let $a_{2}, \cdots, a_{n}$ all be negative, and $a_{1}=-a_{2}-\cdots-a_{n}+1$. At this point, any good permutation $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$ must have $b_{1}=a_{1}$, and $b_{2}, \cdots, b_{n}$ can be any permutation of $a_{2}, \cdots, a_{n}$. Therefore, there are $(n-1)!$ good permutations in this case.

Next, we prove that there are at least $(n-1)!$ good permutations. Note that $(n-1)!$ is the number of different circular permutations of $a_{1}, a_{2}, \cdots, a_{n}$. We first prove that each circular permutation corresponds to a good permutation. For convenience, we call the first term of a good permutation a good number. We only need to prove that each circular permutation must contain a good number.

Method 1: Induction on $n$. When $n=1$, the conclusion is obviously true. Assume that for all $n$, the conclusion holds. For $n+1$, since the sum of all numbers is positive, there is at least one positive number. Group each positive number and the next positive number in the clockwise direction, along with all the numbers between them, into one group. Each group has at least one number, and at least one group has at least two numbers (since not all numbers are positive), so there are at most $k-1$ groups. Summing the numbers in each group, we get fewer than $k$ sums. Write these sums in the order of their groups on the circle. Since the total sum of these sums is positive, by the induction hypothesis, there is a sum that is a good number. Consider the positive number in the group corresponding to this sum; this number is a good number in the entire circular permutation. By the principle of induction, the conclusion holds.

Method 2: Using the extremal principle. For any circular permutation $\left(b_{1}, b_{2}, \cdots, b_{n}\right)$, consider all partial sums starting with $b_{i}$: $b_{i}+b_{i+1}+\cdots+b_{i+t}$, where indices greater than $n$ are taken modulo $n$. For all $i=1,2, \cdots, n$ and all $t=0,1,2, \cdots, n-1$, there must be a smallest partial sum $b_{i}+b_{i+1}+\cdots+b_{i+c}$. Since there is at least one non-positive number, $b_{i}+b_{i+1}+\cdots+b_{i+t} \leqslant 0$. Among all such smallest sums, let the one with the largest number of terms $t+1$ be $b_{i}+b_{i+1}+\cdots+b_{i+t}$. We prove that $b_{i+t+1}$ is a good number.

Indeed, if there exists a positive integer $k$ such that $b_{i+t+1}+b_{i+t+2}+\cdots+b_{i+t+k} \leqslant 0$, then $\left(b_{i}+b_{i+1}+\cdots+b_{i+t}\right)+\left(b_{i+t+1}+b_{i+t+2}+\cdots+b_{i+t+k}\right) \leqslant b_{i}+b_{i+1}+\cdots+b_{i+t}$, which contradicts the fact that $b_{i}+b_{i+1}+\cdots+b_{i+t}$ is the smallest and has the most terms.

Since there are $(n-1)!$ circular permutations, and each circular permutation corresponds to at least one good permutation, and different circular permutations correspond to different good permutations, there are at least $(n-1)!$ good permutations.

In conclusion, the minimum value is $(n-1)!$.