Example 4 Positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (allowing repetition). What is the minimum number of distinct numbers in $a_{2}, \cdots, a_{2006}$? ${ }^{[1]}$
(2006, China Mathematical Olympiad)
This problem is solved using combinatorial geometry methods.
Let the positive integers $a_{1}, a_{2}, \cdots, a_{n}$ be such that $\frac{a_{i}}{a_{i+1}} (1 \leqslant i \leqslant n-1)$ are all distinct. Then the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{n}$ is $k=\left\lceil\frac{1+\sqrt{4 n-7}}{2}\right\rceil$, where $\lceil x\rceil$ denotes the smallest integer not less than the real number $x$.

Since the number of distinct ratios of $k$ distinct positive integers is at most $k(k-1)+1$, we have
$$
k(k-1)+1 \geqslant n-1 \Rightarrow k \geqslant\left\lceil\frac{1+\sqrt{4 n-7}}{2}\right\rceil .
$$

Let $p_{1}, p_{2}, \cdots, p_{k}$ be $k$ distinct prime numbers, and construct a directed graph with these primes as vertices. Connect each pair of distinct vertices $p_{i}$ and $p_{j}$ with two directed edges $p_{i} \rightarrow p_{j}$ and $p_{j} \rightarrow p_{i}$, and add a loop at vertex $p_{1}$. Since this graph is connected and each vertex has equal in-degree and out-degree, it has an Eulerian circuit. Traversing this circuit (which has $k(k-1)+1$ edges) and recording the vertices visited, we obtain a sequence of $k(k-1)+2$ terms that satisfies the given condition (because the prime equality $\frac{p_{i}}{p_{j}}=\frac{p_{k}}{p_{m}}$ holds if and only if $(i, j)=(k, m)$ or $(i, k)=(j, m)$).
Therefore, the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2006}$ is
$$
\lceil\sqrt{2004.25}+0.5\rceil=46
$$