China Mathematical Olympiad

With 45 different positive integers we can only get $45\times 44+1=1981$ fractions. So there are more than 45 different numbers in $\{a_1,a_2,\dots ,a_{2006}\}$.

On the other hand, let $p_1,p_2,\dots ,p_{46}$ be 46 different prime. Set $a_1,a_2,\dots ,a_{2006}$ to be: $$\begin{array}{r}{p}_1,p_1,p_2,p_1,p_3,p_2,p_3,p_1,p_4,p_3,p_4,p_2,p_4,p_1,p_1,p_k,p_{k-1},p_k,p_{k-2},p_k,\dots ,p_k,p_2,p_k,p_1,p_1,p_{45},p_{44},p_{45},p_{43},p_{45},\dots ,p_{45},p_2,p_{45},p_1,p_{46},p_{45},p_{46},p_{44},p_{46},\dots ,p_{46},p_{22},p_{46}.\end{array}$$ Then the 2 006 positive numbers satisfy that any two of $\frac{a_1}{a_2},\frac{a_2}{a_3},\dots ,\frac{a_{2005}}{a_{2006}}$ are unequal. So the answer is 46.