Example 6 Let $n \equiv 1(\bmod 4)$ and $n>1, P=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be any permutation of $\{1,2,3, \cdots, n\}$, and $k_{p}$ denote the largest index $k$ such that the following inequality holds,
$$
a_{1}+a_{2}+\cdots+a_{k}<a_{k+1}+a_{k+2}+\cdots+a_{n} \text {. }
$$

Find the sum of $k_{p}$ for all possible different permutations $P$.
Let $n=4 m+1$ (where $m$ is a positive integer). For any permutation $P=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ of $\{1,2, \cdots, n\}$, by the definition of $k_{p}$, we have
$$
\begin{array}{l}
a_{1}+a_{2}+\cdots+a_{k_{p}}a_{k_{p}+2}+\cdots+a_{n} .
\end{array}
$$

Thus, for the reverse permutation $P^{\prime}=\left\{a_{n}, a_{n-1}, \cdots, a_{2}, a_{1}\right\}$ of $P$, by (1) and (3), we get $k_{p^{\prime}}=n-\left(k_{p}+1\right)$, i.e., $k_{p}+k_{p^{\prime}}=n-1$. Pairing the $n!$ permutations of $\{1,2, \cdots, n\}$ such that each pair consists of a permutation $P$ and its reverse $P^{\prime}$, the sum of the corresponding $k_{p}$ and $k_{p^{\prime}}$ for each pair is $n-1$. Since there are $\frac{n!}{2}$ pairs, the total sum of all $k_{p}$ for all different permutations $P$ is $\frac{1}{2}(n-1)(n!)$.