Assuming positive integer $n$ satisfies $n\equiv 1\pmod{4}$ and $n > 1$. Let $\mathbb{P}=\{a_1,\ a_2,\ \cdots,\ a_n\}$ be a permutation of $\{1,\ 2,\ \cdots,\ n\}$. If $k_p$ denotes the largest index $k$ associated with $\mathbb{P}$ such that the following inequality holds $$a_1 + a_2 + \cdots + a_k < a_{k+1}+a_{k+2}+\cdots + a_n$$
Find the sum of $k_p$ for all possible $\mathbb{P}$.
The given condition implies $$a_1 + a_2 + \cdots + a_{k_p} + a_{k_p+1} \ge a_{k_p+2} + a_{k_p+3}+\cdots + a_n$$
Let's show that the equality cannot hold in the above relation. This is because if $$a_1 + a_2 + \cdots + a_{k_p} + a_{k_p+1} = a_{k_p+2} + a_{k_p+3}+\cdots + a_n$$
Then $$\begin{align*} 2(a_1+a_2+\cdots +a_{k_p+1}) &=\ a_1 + a_2 + \cdots + a_n\\ &=\ 1+2+3+\cdots + n\\&=\ \frac{n(n+1)}{2} \end{align*}$$
Let $n=4m+1$ where $m$ is a positive integer. Then the right side of the above relation equals $(4m+1)(2m+1)$ which is odd. However, the left side of the above relation is even. Therefore, it is impossible. This means $$a_1 + a_2 + \cdots + a_{k_p} + a_{k_p+1} > a_{k_p+2} + a_{k_p+3}+\cdots + a_n$$
It follows for any $\mathbb{P}=\{a_1,\ a_2,\ \cdots,\ a_n\}$ with $$a_1 + a_2 + \cdots + a_{k_p} < a_{k_p+1}+a_{k_p+2}+\cdots + a_n$$
its reversed permutation $\mathbb{P'}=\{a_n,\ a_{n-1},\ \cdots,\ a_1\}$ must satisfy $$\underbrace{a_n + \cdots + a_{k_p+2}}_{n-k_p-1} < a_{k_p+1} +a_{k_p}+ \cdots + a_{1}$$
and $$a_n + \cdots + a_{k_p+2} + a_{k_p+1} > a_{k_p} + \cdots + a_{1}$$
In other words, the largest indexes of $\mathbb{P}$ and $\mathbb{P'}$ satisfying the given requirement must be $k_p$ and $(n-k_p-1)$ whose sum is $(n-1)$. There are totally $n!$ different permutation. Hence, the answer is $$\boxed{\frac{(n-1)(n!)}{2}}$$