(5) Let $a_{1}, a_{2}, \cdots, a_{10}$ be 10 distinct natural numbers, whose sum is 1995. Try to find the minimum value of
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{9} a_{10}+a_{10} a_{1}
$$
Given $\sum_{i=1}^{10} a_{i}=1995$ and $a_{1}, a_{2}, \cdots, a_{10}$ are natural numbers, the minimum value exists.

Let $S=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{10} a_{1}$ be the minimum value. Without loss of generality, assume $a_{1}$ is the largest among $a_{1}, a_{2}, \cdots, a_{10}$.
$a_{2}+a_{10}$ should not exceed $a_{k-1}+a_{k+1}\left(k=3,4, \cdots, 9\right.$, with the convention $\left.a_{11}=a_{1}\right)$, otherwise, swapping $a_{1}$ and $a_{k}$ would result in:
$$
\begin{array}{l}
a_{1}\left(a_{2}+a_{10}\right)+a_{k}\left(a_{k-1}+a_{k+1}\right)> \\
a_{k}\left(a_{2}+a_{10}\right)+a_{1}\left(a_{k-1}+a_{k+1}\right)
\end{array}
$$
$S=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{10} a_{1}$ would be smaller after the adjustment.
If $\left\{a_{2}, a_{3}, \cdots, a_{10}\right\} \neq\{1,2, \cdots, 9\}$, then there must be some $a_{k}>j, k \in$ $\{2,3, \cdots, 10\}, j \in\{1,2, \cdots, 9\}$, and $a_{k} \notin\{1,2, \cdots, 9\}, j \notin\left\{a_{2}\right.$, $\left.a_{3}, \cdots, a_{10}\right\}$.
When $k \neq 2,10$, changing $a_{k}$ to $j$ and $a_{1}$ to $a_{1}+a_{k}-j$ results in:
$$
\begin{array}{l}
\left(a_{1}+a_{k}-j\right)\left(a_{2}+a_{10}\right)+j\left(a_{k-1}+a_{k+1}\right)< \\
a_{1}\left(a_{2}+a_{10}\right)+a_{k}\left(a_{k-1}+a_{k+1}\right)
\end{array}
$$
$S$ would be smaller after the adjustment. Therefore, when $S$ is minimized, $a_{3}, a_{4}, \cdots, a_{9} \in\{1,2, \cdots, 9\}$. Since $a_{2}+a_{10}<a_{2}+a_{4}$ and $a_{2}+a_{10}<a_{8}+a_{10}$, it follows that $a_{2}<a_{8}$ and $a_{10}<a_{4}$, hence
$$
\left\{a_{2}, a_{3}, \cdots, a_{10}\right\}=\{1,2, \cdots, 9\}
$$

Thus, $a_{1}=1995-\sum_{i=2}^{10} i=1995-45=1950$
Therefore, by the minimality of $S$, we must have
$$
a_{2}+a_{10}=1+2
$$

Assume $a_{2}=1, a_{10}=2$, then
$$
S=3 \times 1950+a_{3}+a_{3} a_{4}+a_{4} a_{5}+\cdots+a_{7} a_{8}+a_{8} a_{9}+2 a_{9}
$$

Then
$$
\begin{array}{l}
2 a_{3}+a_{3} a_{4}+a_{4} a_{5}+\cdots+a_{7} a_{8}+a_{8} a_{9}+2 a_{9}-a_{3}= \\
-\frac{1}{2}\left(\left(2-a_{3}\right)^{2}+\left(a_{3}-a_{4}\right)^{2}+\cdots+\left(a_{9}-2\right)^{2}\right]+ \\
2^{2}+3^{2}+\cdots+9^{2}-a_{3} \geqslant
\end{array}
$$
$$
\begin{array}{l}
-\frac{1}{2}\left((2-9)^{2}+(9-3)^{2}+(8-2)^{2}+(8-4)^{2}+(7-3)^{2}+\right. \\
\left.(7-5)^{2}+(6-4)^{2}+(6-5)^{2}\right)+2^{2}+3^{2}+\cdots+9^{2}-9= \\
9 \times(1+3)+8 \times(2+4)+7 \times(3+5)+6 \times(4+5)
\end{array}
$$

Thus, when $a_{1}=1950, a_{2}=1, a_{10}=2, a_{3}=9, a_{4}=3, a_{9}=8$,
$$
\begin{aligned}
a_{5}=7, a_{8}= & 4, a_{6}=5, a_{7}=6 \text {, } S \text { achieves its minimum value } \\
& 3 \times 1950+9 \times(1+3)+8 \times(2+4)+ \\
& 7 \times(3+5)+6 \times(4+5)=6044
\end{aligned}
$$