9 Let $a_{1}, a_{2}, \cdots, a_{10}$ be 10 distinct positive integers, with a sum of 2002. Find the minimum value of $a_{1} a_{2}+$ $a_{2} a_{3}+\cdots+a_{10} a_{1}$.
9. The solution to this problem is carried out in two steps:
(1) First, consider how to arrange $a_{1}, a_{2}, \cdots, a_{10}$ to minimize the corresponding sum when $a_{1}, a_{2}, \cdots, a_{10}$ are determined. First, observe a simple case, i.e., $a_{1}+a_{2}+\cdots+a_{10}=55$, $\left\{a_{1}, a_{2}, \cdots, a_{10}\right\}=\{1,2, \cdots, 10\}$. Arrange $a_{1}, a_{2}, \cdots, a_{10}$ in a circle, and we prove: 10 must be adjacent to 1 and 2.

If 1 is not adjacent to 10, then flip the segment $a_{i}, a_{i+1}, \cdots, 1$ as a whole. Let $S=a_{1} a_{2} + a_{2} a_{3} + \cdots + a_{10} a_{1}$, and $S^{\prime}$ be the corresponding sum after the flip. Then,
$$S^{\prime}-S=10 \cdot 1 + a_{i} \cdot a_{j} - \left(10 \cdot a_{i} + a_{j} \cdot 1\right) \leqslant 0,$$

Thus, the sum does not increase after the adjustment. Similarly, it can be proven that the other side of 10 must be 2.
Using the same method, it can be proven that the other side of 1 must be 9.
By analogy, the arrangement that minimizes $S$ can only be $10,1,9,3,7,5,6,4,8,2$ (circular arrangement), at which point $S_{\text {min }}=224$.
(2) Next, consider the case where $a_{1}+a_{2}+\cdots+a_{10}=2002$.
Let $\operatorname{minS}(n)=g(n)$ when $a_{1}+a_{2}+\cdots+a_{10}=n$, then $g(55)=224$.
We now prove: $g(n+1) \geqslant g(n)+3$.
For any arrangement $a_{1}, a_{2}, \cdots, a_{10}$ (in a circle) where $a_{1}+a_{2}+\cdots+a_{10}=n+1$.
If $a_{1}, a_{2}, \cdots, a_{10}$ are 10 consecutive positive integers, assume $a_{1}>1$ (if $a_{1}=1$, then $n=55$), then $a_{1}-1$ is not among the original 10 numbers. Replace $a_{1}$ with $a_{1}-1$, and the sum decreases by at least $a_{1}(1+2) - (a_{1}-1)(1+2)=3$.

If $a_{1}, a_{2}, \cdots, a_{10}$ are not consecutive positive integers, then there must be an $a_{i}$ such that $a_{i}-1$ is not among the original 10 numbers. Replace $a_{i}$ with $a_{i}-1$, and the sum also decreases by at least 3. Therefore, for every arrangement $S(n+1)$ of $a_{1}+a_{2}+\cdots+a_{10}=n+1$, there exists an arrangement $a_{1}, a_{2}, \cdots, a_{i}-1, \cdots, a_{10}$ such that $S(n+1) \geqslant S(n)+3 \geqslant g(n)+3$, hence $\min S(n+1) \geqslant g(n)+3$, i.e., $g(n+1) \geqslant g(n)+3$, thus $g(2002) \geqslant 6065$, and the equality can be achieved: $a_{1}=1, a_{2}=9, a_{3}=3, a_{4}=7, a_{5}=5, a_{6}=6, a_{7}=4, a_{8}=8, a_{9}=2, a_{10}=1957$.