6. 121 Let $a_{1}, a_{2}, \cdots, a_{10}$ be any 10 distinct natural numbers, and their sum is 1995. Try to find the minimum value of
$$a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{9} a_{10}+a_{10} a_{1}$$
[Solution] Write 10 natural numbers in a clockwise direction on a circle. Thus, the expression in the problem is the sum of the products of every two adjacent numbers, which we will simply call the "sum value".

Let the sum of 10 distinct natural numbers be $N$, then $N \geqslant 55$. Denote the minimum "sum value" of any 10 distinct natural numbers with a sum of $N$ as $S(N)$.

First, consider the case when $N=55$.
Obviously, at this time $\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}=\{1,2, \cdots, 10\}$.
We will prove that when the 10 numbers are arranged as in Figure 1 or Figure 2, the "sum value" is the smallest.
Assume these 10 numbers are arranged arbitrarily on the circle and are different from Figure 1 and Figure 2. Without loss of generality, let $a_{1}=10$ (as shown in Figure 3). If $a_{2} \neq 1$ and $a_{j}=1, j \neq 2$. We will reverse the order of the segment $\left(a_{2}, a_{3}, \cdots, a_{j}\right)$ to $\left(a_{j}, a_{j-1}, \cdots, a_{2}\right)$, keeping the rest of the numbers unchanged, resulting in Figure 4, and call this process one operation. The change in the "sum value" before and after the operation is
$$10 a_{2}+a_{j+1}-\left(10+a_{2} a_{j+1}\right)=\left(a_{2}-1\right)\left(10-a_{j+1}\right) \geqslant 0$$
The "sum value" remains unchanged only when $a_{j}=a_{10}=1$, otherwise it decreases.
Relabel the letters in Figure 4 in a clockwise direction starting from $a_{1}$ as $b_{1}, b_{2}, \cdots, b_{10}$. Then $b_{1}=10, b_{2}=1$. If $b_{10} \neq 2, b_{j}=2,3 \leqslant j \leqslant 9$ (as shown in Figure 5), we will reverse the order of the segment $\left(b_{j}, b_{j+1}, \cdots, b_{10}\right)$ to $\left(b_{10}, \cdots b_{j+1}, b_{j}\right)$, keeping the rest of the numbers unchanged, resulting in Figure 6. The change in the "sum value" before and after this operation is
$$\begin{array}{l}
10 b_{10}+2 b_{j-1}-\left(2 \times 10+b_{10} \cdot b_{j-1}\right) \\
=\left(b_{10}-2\right)\left(10-b_{j-1}\right)>0
\end{array}$$

It is evident that the "sum value" decreases.
If Figure 6 is still different from Figure 2, then we can use a similar operation to transform it step by step into Figure 2, and each step will decrease the "sum value".

Notice that the "sum value" corresponding to the arrangement in Figure 1 and Figure 2 is the same. Therefore, when $N=55$, the "sum value" is the smallest only when the 10 numbers are arranged as in Figure 1 or Figure 2. Thus,
$$\begin{aligned}
S(55)= & 10 \times 1+1 \times 9+9 \times 3+3 \times 7+7 \times 5+5 \times 6+6 \times 4+4 \\
& \times 8+8 \times 2+2 \times 10=224
\end{aligned}$$

Next, consider the case when $N>55$.
At this time, the 10 distinct natural numbers with a sum of $N$ can not only be arranged in different orders but also in different ways of partitioning $N$. Therefore, when discussing the minimum "sum value", we must examine all possible partitioning methods of $N$ and all different arrangement forms.

For any partitioning and any arrangement on the circle of the 10 numbers, we use $a_{1}$ to represent the largest of the 10 numbers, and label the 10 numbers in a clockwise direction starting from $a_{1}$ as $a_{1}, a_{2}, \cdots, a_{10}$. We use
$$N=a_{1}+a_{2}, \cdots+a_{10}$$

not only to indicate the equality but also to indicate that the 10 distinct natural numbers with $a_{1}$ as the largest are arranged in the order mentioned above on the circle.

Since $N>55$, when these 10 numbers are 10 consecutive natural numbers, the smallest number $a_{i_{0}}>1$; and when these 10 numbers are not 10 consecutive natural numbers, there must be a number $a_{i_{0}}$ that is at least 2 greater than all numbers smaller than it. Thus, we obtain a partitioning and arrangement method for $N-1$
$$N-1=a_{1}+\cdots+a_{i_{0}-1}+\left(a_{i_{0}}-1\right)+a_{i_{0}+1}+\cdots+a_{10}$$
where $a_{0}=a_{10}$. This means that for every (1) equation, there is at least one (2) equation corresponding to it. Therefore,
$$\begin{aligned}
S(N)-S(N-1) & \geqslant \min \left(a_{i_{0}-1}+a_{i_{0}+1}\right) \\
& \geqslant 2+1=3
\end{aligned}$$

Thus, we have
$$\begin{aligned}
S(N) & \geqslant S(55)+3(N-55) \\
& =3 N+79
\end{aligned}$$

On the other hand, when
$$\left(a_{1}, a_{2}, \cdots, a_{10}\right)=(N-45,1,9,3,7,6,5,4,8,2)$$

the "sum value" is indeed $3 N+79$.
Therefore, $S(N)=3 N+79$.
In particular,
$$S(1995)=3 \times 1995+79=6064$$