2・66 Let $\left[a_{1}, a_{2}, \cdots, a_{n}\right]$ be a permutation of $1,2, \cdots, n$. If $a_{i}>a_{j}$ for all $j>i$, then $a_{i}$ is called a “large” number in this permutation.
Try to calculate the arithmetic mean of all “large” numbers in all $n$! permutations.
[Solution] Let $\sigma$ be a permutation, and denote $N_{i}(\sigma)=1$ if the $i$-th number in $\sigma$ is a “large” number, and $N_{i}(\sigma)=0$ if the $i$-th number in $\sigma$ is not a “large” number.

It can be proven that among all $n!$ permutations from 1 to $n$, the average value of $N_{i}(\sigma)$ is $\frac{1}{n-i+1}$.

This is because the $i-1$ numbers before $a_{i}$ have $P_{n}^{i-1}$ ways of arrangement, and then one of the largest numbers from the remaining $n-i+1$ numbers is placed at the $i$-th position, with the arrangements for the positions after the $i+1$-th position being $P_{n-i}^{n-i}$. Therefore, we have
$$
\sum N_{i}(\sigma)=P_{n}^{n-1} \cdot P_{n-i}^{n-i}=n(n-1) \cdots(n-i+2)(n-i)!
$$

Thus, the average value of $N_{i}(\sigma)$ is
$$
\frac{n(n-1) \cdots(n-i+2)(n-i)!}{n!}=\frac{1}{n-i+1} \text {. }
$$

Taking $i=1,2, \cdots, n$, we obtain the average value of “large” numbers in this problem as
$$
\frac{1}{n}+\frac{1}{n-1}+\frac{1}{n-2}+\cdots+\frac{1}{2}+1 .
$$