18. (2 marks) Let $A_{1} A_{2} \cdots A_{2002}$ be a regular 2002-sided polygon. Each vertex $A_{i}$ is associated with a positive integer $a_{i}$ such that the following condition is satisfied: If $j_{1}, j_{2}, \cdots, j_{k}$ are positive integers such that $k<500$ and $A_{j_{1}} A_{j_{2}} \cdots A_{j_{k}}$ is a regular $k$-sided polygon, then the values of $a_{j_{1}}, a_{j_{2}}, \cdots, a_{j_{k}}$ are all different. Find the smallest possible value of $a_{1}+a_{2}+\cdots+a_{2002}$.
(2 分) 設 $A_{1} A_{2} \ldots A_{2002}$ 為正 2002 邊形 0 每一頂點 $A_{i}$ 對應正整數 $a_{i}$ 使下列條件成立:如果 $j_{1}, j_{2}, \cdots, j_{k}$為正整數, $k<500$, 而 $A_{j_{1}} A_{j_{2}} \ldots A_{j_{k}}$ 為正 $k$ 邊形時, 則 $a_{j_{1}}, a_{j_{2}}, \cdots, a_{j_{k}}$ 的值互不相同。求 $a_{1}+a_{2}+\ldots+a_{2002}$的最小可能值。
18. Since $2002=2 \times 7 \times 11 \times 13$, we can choose certain vertices among the 2002 vertices to form regular 7-, 11-, 13-, $\cdots$ polygons (where the number of sides runs through divisors of 2002 greater than 2 and less than 1001). Since $2002=7 \times 286$, so there are at least 286 different positive integers among the $a_{i}{ }^{\prime} s$. Hence the answer is at least $7 \times(1+2+\cdots+286)=\frac{7 \times 286 \times 287}{2}=287 \times 1001=287287$. Indeed this is possible if we assign the number $\left[\frac{i}{7}\right]+1$ to each $A_{i}$, for $i=1,2, \cdots, 2002$.