With 45 different positive integers we can only get $45 \times 44 + 1 = 1\,981$ fractions. So there are more than 45 different numbers in $\{a_1, a_2, \dots, a_{2\,006}\}$.

On the other hand, let $p_1, p_2, \dots, p_{46}$ be 46 different prime. Set $a_1, a_2, \dots, a_{2\,006}$ to be:
$$\begin{align*} p_1, p_1, p_2, p_1, p_3, p_2, p_3, p_1, p_4, p_3, p_4, p_2, p_4, p_1, \ p_1, p_k, p_{k-1}, p_k, p_{k-2}, p_k, \dots, p_k, p_2, p_k, p_1, \ p_1, p_{45}, p_{44}, p_{45}, p_{43}, p_{45}, \dots, p_{45}, p_2, p_{45}, p_1, \ p_{46}, p_{45}, p_{46}, p_{44}, p_{46}, \dots, p_{46}, p_{22}, p_{46}. \end{align*}$$
Then the 2 006 positive numbers satisfy that any two of $\frac{a_1}{a_2}, \frac{a_2}{a_3}, \dots, \frac{a_{2\,005}}{a_{2\,006}}$ are unequal.
So the answer is 46.