15. (30 points) Let $\sigma=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ be a permutation of $\{1$, $2, \cdots, n\}$, and define $F(\sigma)=\sum_{i=1}^{n} a_{i} a_{i+1}$, with $a_{n+1}=a_{1}$. Find $\min F(\sigma)$.
15. The problem is equivalent to placing $n$ numbers on a circle so that the sum of the products of adjacent numbers is minimized, with the minimum value denoted as $T_{n}$.

Assume $a_{1}=n$, then the number 1 must be adjacent to it. Otherwise, let $a_{j}=1(j \notin\{2, n\})$. We can change the numbers $a_{2}, a_{3}, \cdots, a_{j}$ to the numbers on $a_{i}, a_{j-1}, \cdots, a_{2}$, then the change in the sum of the products of adjacent numbers is
$$
\begin{array}{c}
a_{1} a_{j}+a_{2} a_{j+1}-a_{1} a_{2}-a_{j} a_{j+1} \\
=\left(a_{1}-a_{j+1}\right)\left(a_{j}-a_{2}\right)<0 .
\end{array}
$$

Thus, we can determine $a_{2}=1$.
The number 2 must also be adjacent to the number $n$, i.e., $a_{n}=2$.
In fact, if $a_{j}=2(j \neq n)$, then swap $a_{n}$, $a_{n-1}, \cdots, a_{j}$ to $a_{j}, a_{j+1}, \cdots, a_{n}$. At this time, the change in the target value is
$$
\begin{array}{c}
a_{1} a_{j}+a_{n} a_{j-1}-a_{1} a_{n}-a_{j} a_{j-1} \\
=\left(a_{1}-a_{j-1}\right)\left(a_{j}-a_{n}\right)<0 .
\end{array}
$$

Therefore, when the target value is minimized,
$$
a_{1}=n, a_{2}=1, a_{n}=2 \text {. }
$$

From this, we can sequentially get $a_{3}=n-1, a_{n-1}=n-2$.
Now that the numbers at both ends have been arranged, if the remaining numbers are all smaller than the numbers at both ends, then find the two smallest numbers among the remaining numbers and place them in the order of small to large and large to small; if the remaining numbers are all larger than the numbers at both ends, then find the two largest numbers among the remaining numbers and place them in the order of large to small and small to large. By this rule, we get
$$
a_{4}=3, a_{n-2}=4, a_{5}=n-3, a_{n-3}=n-4, \cdots
$$

Below, we use the recursive method to calculate $T_{n}$.
Consider $n+2$ numbers. In the number sequence of $T_{n}$, add 1 to each number, then place the numbers 1 and $n+2$ in the middle of 2 and $n+1$, i.e., we get $T_{n+2}$.
$$
\begin{array}{l}
\text { Then } T_{n+2} \\
=T_{n}^{\prime}+(n+1)+(n+2)+2(n+2)-2(n+1) \text {, }
\end{array}
$$

where, $T_{n}^{\prime}=\sum_{i=1}^{n}\left(a_{i}+1\right)\left(a_{i+1}+1\right)$
$$
=T_{n}+n(n+2) \text {. }
$$

Thus, $T_{n+2}=T_{n}+n^{2}+4 n+5$
$$
\Rightarrow T_{n}=\left\{\begin{array}{ll}
\frac{1}{6} n^{3}+\frac{1}{2} n^{2}+\frac{5}{6} n-1, & n=2 m ; \\
\frac{1}{6} n^{3}+\frac{1}{2} n^{2}+\frac{5}{6} n-\frac{1}{2}, & n=2 m-1 .
\end{array}\right.
$$
(Yang Xiaoming provided)