2 Positive integers $a_{1}, a_{2}, \cdots, a_{2006}$ (which can be the same) 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 $a_{1}, a_{2}, \cdots, a_{2006}$? (Provided by Yonggao Chen)
Solution: Among $a_{1}, a_{2}, \cdots, a_{2006}$, there are at least 46 distinct numbers.
Since the pairwise ratios of 45 distinct positive integers can have at most $45 \times 44 + 1 = 1981$ values, the number of distinct numbers in $a_{1}, a_{2}, \cdots, a_{2006}$ must be greater than 45.
Below is an example to show that 46 is achievable.
Let $p_{1}, p_{2}, \cdots, p_{46}$ be 46 distinct prime numbers, and construct $a_{1}, a_{2}, \cdots, a_{2006}$ as follows:
$$
\begin{array}{c}
p_{1}, p_{1}, p_{2}, p_{1}, p_{3}, p_{2}, p_{3}, p_{1}, p_{4}, p_{3}, p_{4}, p_{2}, p_{4}, p_{1}, \cdots, \\
p_{1}, p_{k}, p_{k-1}, p_{k}, p_{k-2}, p_{k}, \cdots, p_{k}, p_{2}, p_{k}, p_{1}, \cdots, \\
p_{1}, p_{45}, p_{44}, p_{45}, p_{43}, p_{45}, \cdots, p_{45}, p_{2}, p_{45}, p_{1}, \\
p_{46}, p_{45}, p_{46}, p_{44}, p_{46}, \cdots, p_{46}, p_{22}, p_{46},
\end{array}
$$

These 2006 positive integers meet the requirements.
Therefore, $a_{1}, a_{2}, \cdots, a_{2006}$ must have at least 46 distinct numbers.