2.70 Let $a_{1}, a_{2}, \cdots, a_{10}$ be any 10 pairwise distinct natural numbers, and $a_{1}+a_{2}+\cdots+a_{10}=1995$. Try to find the minimum value of
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{9} a_{10}+a_{10} a_{1}
$$
[Solution] Given 10 natural numbers $a_{1}, a_{2}, \cdots, a_{10}$, write them in a clockwise direction on a circle. The expression in question is the sum of the products of each pair of adjacent integers, denoted as $S(m)$, where $m=a_{1}+a_{2}+\cdots+a_{10}$.

Obviously, the minimum sum of 10 distinct natural numbers is 55. Therefore, we first examine the case when $m=55$. In this case, $\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}=\{1,2, \cdots, 10\}$. Let's prove that when
$$
\begin{array}{l}
a_{1}=10, a_{2}=1, a_{3}=9, a_{4}=3, a_{5}=7, \\
a_{6}=5, a_{7}=6, a_{8}=4, a_{9}=8, a_{10}=2 \text {, }
\end{array}
$$
$S(55)$ is minimized.
For any arrangement of these 10 natural numbers on the circle, assume $a_{1}=10$. If $a_{2} \neq 1$, let $a_{j}=1, 3 \leqslant j \leqslant 10$. In this case, we reverse the order of the numbers on the arc segment $a_{2} a_{j}$, i.e., change $\left\{a_{2}, a_{3}, \cdots, a_{j-1}, a_{j}=1\right\}$ to $\left\{1, a_{j-1}\right.$, $\left.\cdots, a_{3}, a_{2}\right\}$. This operation changes the sum $S$ only in two places, and the change in value is
$$
\begin{array}{l}
10 a_{2}+1 \cdot a_{j+1}-\left(1 \cdot 10+a_{2} a_{j+1}\right) \\
=\left(a_{2}-1\right)\left(10-a_{j+1}\right) \geqslant 0 .
\end{array}
$$

This shows that during this operation, the sum $S$ does not increase, but after the operation, we have $a_{1}=10, a_{2}=1$.

If $a_{10} \neq 2$, let $a_{i}=2, 3 \leqslant i \leqslant 9$. Similarly, reverse the order of the numbers on the arc segment $a_{i} a_{10}$, i.e., change $\left\{a_{i}, a_{i+1}, \cdots, a_{10}\right\}$ to $\left\{a_{10}, a_{9}, \cdots, a_{i+1}\right.$, $\left.a_{i}\right\}$. The change in the sum $S$ is
$$
\begin{aligned}
& a_{i-1} a_{i}+a_{10} a_{1}-\left(a_{i} a_{1}+a_{i-1} a_{10}\right) \\
= & \left(10-a_{i-1}\right)\left(a_{10}-2\right)>0,
\end{aligned}
$$

which means the sum $S$ decreases after the operation.
If $a_{3} \neq 9$, we can similarly perform the operation to make $a_{3}=9$ and decrease the sum $S$. Continuing this process, we can achieve the desired result.
Thus, the minimum value of $S(55)$ is
$$
\begin{aligned}
S_{0}(55)= & 10 \times(1+2)+9 \times(1+3)+8 \times(2+4) \\
& +7 \times(3+5)+6 \times(4+5)=224 .
\end{aligned}
$$

Next, we examine the relationship between $S(k)$ and $S(k-1)$ for $k>55$. Arrange the 10 natural numbers with a sum of $k$ in descending order as
$$
b_{1}>b_{2}>b_{3}>\cdots>b_{10} .
$$

Let $b_{11}=0$ and
$$
d_{i}=b_{i}-b_{i+1}, i=1,2, \cdots, 10,
$$

then $d_{i} \geqslant 1$. Since $b_{1}+b_{2}+\cdots+b_{10}=k>55$, at least one of $\left\{d_{1}, d_{2}, \cdots, d_{10}\right\}$ is greater than 1. Subtract 1 from the number represented by $b_{i}$, then the 10 natural numbers are still distinct and their sum is $k-1$. In this case, it is clear that
$$
S(k) \geqslant S(k-1)+3 .
$$

Since the above inequality holds for any arrangement, it also holds for the minimum value, i.e.,
$$
S_{0}(k) \geqslant S_{0}(k-1)+3 \text {. }
$$

From (1) and (2), we have
$$
\begin{aligned}
S_{0}(1995) & \geqslant S_{0}(55)+3 \times 1940 \\
& =6044 .
\end{aligned}
$$

On the other hand, when
$$
\begin{array}{l}
a_{1}=1950, a_{2}=1, a_{3}=9, a_{4}=3, a_{5}=7, \\
a_{6}=5, a_{7}=6, a_{8}=4, a_{9}=8, a_{10}=2
\end{array}
$$

we have
$$
S(1995)=6044 .
$$

From (3) and (4), the minimum value is 6044.