20. Let $a_{1}, a_{2}, \cdots, a_{10}$ be any ten distinct natural numbers, whose sum is 1995. Try to find the minimum value of $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{9} a_{10}+a_{10} a_{1}$.
20. Given that $a_{1}, a_{2}, \cdots, a_{10}$ have only a finite number of possibilities, the minimum value exists.

Let $S=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{10} a_{1}$, which is already the minimum. Without loss of generality, assume $a_{1}, a_{2}, \cdots, a_{10}$ where $a_{1}$ is the largest. $a_{2}+a_{10}$ should not be greater than $a_{k-1}+a_{k+1}\left(k=3,4, \cdots, 9\right.$, with the convention $\left.a_{11}=a_{1}\right)$, otherwise, swap $a_{1}$ and $a_{k}$.

Since $a_{1}\left(a_{2}+a_{10}\right)+a_{k}\left(a_{k-1}+a_{k+1}\right)>a_{k}\left(a_{2}+a_{10}\right)+a_{1}\left(a_{k-1}+a_{k+1}\right)$, $S=a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{10} a_{1}$ becomes smaller after adjustment.

If $\left\{a_{2}, a_{3}, \cdots, a_{10}\right\} \neq\{1,2, \cdots, 9\}$, then there must be some $a_{k}>j, k \in\{2,3, \cdots, 10\}, j \in\{1,2, \cdots, 9\}$, i.e., $a_{k} \in\{1,2, \cdots, 9\}, j \notin\left\{a_{2}, a_{3}, \cdots, a_{10}\right\}$,
When $k \neq 2,10$, change $a_{k}$ to $j, a_{1}$ to $a_{1}+a_{k}-j$, since
$\left(a_{1}+a_{k}-j\right)\left(a_{2}+a_{10}\right)+j\left(a_{k-1}+a_{k+1}<a_{1}\left(a_{2}+a_{10}\right)+a_{k}\left(a_{k-1}+a_{k+1}\right), S\right.$ becomes smaller after adjustment. Therefore, when $S$ is already the minimum, $a_{3}, a_{4}, \cdots, a_{9} \in\{1,2, \cdots, 9\}$,
Furthermore, by $a_{2}+a_{10}<a_{2}+a_{4}$ and $a_{8}+a_{10}$, we get $a_{2}<a_{8}, a_{10}<a_{4}$, so $\left\{a_{2}, a_{3}, \cdots, a_{10}\right\}=\{1,2, \cdots, 9\}$, thus $a_{1}=1950$.
Then, by the minimality of $S$, we must have $a_{2}+a_{10}=1+2$ (the sum of the eight terms excluding $a_{1}$ is obviously $<$ 1950). Without loss of generality, let $a_{2}=1, a_{10}=2$,
$$
S=3 \times 1950+a_{3}+a_{3} a_{4}+a_{4} a_{5}+a_{6} a_{7}+a_{7} a_{8}+a_{8} a_{9}+2 a_{9},
$$

and $2 a_{3}+a_{3} a_{4}+a_{4} a_{5}+\cdots+a_{8} a_{9}+2 a_{9}-a_{3}$
$$
\begin{aligned}
= & -\frac{1}{2}\left[\left(2-a_{3}\right)^{2}+\left(a_{3}-a_{4}\right)^{2}+\cdots+\left(a_{9}-2\right)^{2}\right]+2^{2}+3^{2}+\cdots+9^{2}-a_{3} \\
\geqslant & -\frac{1}{2}\left[(2-9)^{2}+(9-3)^{2}+(8-2)^{2}+(8-4)^{2}+(7-3)^{2}+(7-5)^{2}+(6\right. \\
& \left.-4)^{2}+(6-5)^{2}\right]+2^{2}+3^{2}+\cdots+9^{2}-9 \\
= & 9(1+3)+8(2+4)+7(3+5)+6(4+5) .
\end{aligned}
$$

Thus, when $a_{1}=1950, a_{2}=1, a_{10}=2, a_{3}=9, a_{4}=3, a_{9}=8, a_{5}=7, a_{8}=4, a_{6}=$ $5, a_{7}=6$, $S$ achieves its minimum value, which is 6044.