Example 4 Let $a_{1}, a_{2}, \cdots, a_{10}$ be any 10 distinct positive integers, and $a_{1}+a_{2}+\cdots+a_{10}=1995$. Try to find the minimum value of $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{10} a_{1}$.
(1995, China Mathematical Olympiad)
[Analysis] Analyze from small to large, find the pattern.
Solve: Write $a_{1}, a_{2}, \cdots, a_{10}$ in a clockwise direction on a circle, and let the expression (i.e., the sum of the products of each pair of adjacent numbers) be $S_{(m)}$, where $m=a_{1}+a_{2}+\cdots+a_{10}$.
Obviously, $m_{\min }=55$.
Consider $m=55$, at this time,
$$
\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}=\{1,2, \cdots, 10\} \text {. }
$$

First, prove that when
$$
\begin{array}{l}
a_{1}=10, a_{2}=1, a_{3}=9, a_{4}=3, a_{5}=7, \\
a_{6}=5, a_{7}=6, a_{8}=4, a_{9}=8, a_{10}=2
\end{array}
$$

$S_{(55) \min }=224$.
Thus, for any permutation of these 10 natural numbers, assume $a_{1}=10, a_{2} \neq 1, a_{j}=1(3 \leqslant j \leqslant 10)$, at this time, reverse the order of all integers on the arc $\widetilde{a}_{2} a_{j}$, i.e., $\left\{a_{2}, a_{3} \cdots, a_{j}\right\}$ is rearranged as $\left\{a_{j}, a_{j-1}, \cdots, a_{2}\right\}$. Then
$$
\begin{array}{l}
S_{55}^{\prime}-S_{(55)} \\
=10 a_{2}+1 \cdot a_{j+1}-\left(1 \times 10+a_{2} a_{j+1}\right) \\
=\left(a_{2}-1\right)\left(10-a_{j+1}\right) \geqslant 0 .
\end{array}
$$

Thus, to minimize $S$, $a_{1}=10, a_{2}=1$.
Similarly, when $a_{10} \neq 2$, assume $a_{j}=2(3 \leqslant j \leqslant 9)$, and repeat the above steps to prove that the values of $a_{1}$ to $a_{10}$ given are the values that minimize $S_{(55)}$.

Next, consider the relationship between $S_{(k)}(k>55)$ and $S_{(k-1)}$.
Arrange the 10 positive integers whose sum is $k$ as
$$
\begin{array}{l}
b_{1}>b_{2}>\cdots>b_{10} . \\
\text { Let } b_{11}=0, d_{i}=b_{i}-b_{i+1}\left(i=1,2, \cdots, 10, d_{i} \geqslant 1\right) .
\end{array}
$$

Since $b_{1}+b_{2}+\cdots+b_{10}=k>55$, there exists $i(1 \leqslant i \leqslant 10)$ such that $d_{i}>1$.

Subtract 1 from the $a_{j}$ represented by $b_{i}$, then the 10 positive integers are all different, and their sum is $k-1$. At this time,
$$
\begin{array}{l}
S_{(k)}-S_{(k-1)} \geqslant 3 . \\
\text { Hence } S_{(1995)}-S_{(55)} \\
\geqslant 3 \times 1940+S_{(55) \min }=6044 . \\
\text { When } a_{1}=1950, a_{3}=9, a_{4}=3, a_{2}=1, \\
a_{5}=7, a_{6}=5, a_{7}=6, a_{8}=4, a_{9}=8, \\
a_{10}=2
\end{array}
$$

$S_{(1995)}=6044$ can be achieved.
Therefore, the minimum value of $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{10} a_{1}$ is 6044.