Two positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (which can be the same) are such that $\frac{a_{1}}{a_{2}}, \frac{a_{2}}{a_{3}}, \cdots, \frac{a_{2005}}{a_{2006}}$ are all distinct. How many different numbers are there at least among $a_{1}$, $a_{2}, \cdots, a_{2006}$?
（Chen Yonggao, problem contributor）
Due to the fact that the pairwise ratios of 45 distinct positive integers are at most $45 \times 44 + 1 = 1981$, the number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2000}$ is greater than 45.
Below is an example to show that 46 can be achieved.
Let $p_{1}, p_{2}, \cdots, p_{46}$ be 46 distinct prime numbers, and construct $a_{1}, a_{2}, \cdots, a_{2000}$ as follows:
$$
\begin{array}{l}
p_{1}, p_{1}, \\
p_{2}, p_{1}, \\
p_{3}, p_{2}, p_{3}, p_{1}, \\
p_{4}, p_{3}, p_{4}, p_{2}, p_{4}, p_{1}, \\
\ldots \\
p_{k}, p_{k-1}, p_{k}, p_{k-2}, \cdots, p_{k}, p_{2}, p_{k}, p_{1}, \\
\ldots \\
p_{45}, p_{44}, p_{45}, p_{43}, \cdots, p_{45}, p_{2}, p_{45}, p_{1}, \\
p_{46}, p_{45}, p_{46}, p_{44}, \cdots, p_{46}, p_{34} .
\end{array}
$$

These 2006 positive integers meet the requirements.
Therefore, the minimum number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2000}$ is 46.