Solution:

The desired value is $C=\frac{1}{2}$.

First we prove that $C \leq \frac{1}{2}$ holds. To do this we assume without loss of generality that $a_{1} \leq a_{2} \leq a_{3} \leq a_{4} \leq a_{5}$ and consider the five fractions $\frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, \frac{a_{1}}{a_{5}}, \frac{a_{2}}{a_{3}}, \frac{a_{4}}{a_{5}}$. By the pigeonhole principle, three distinct ones of these fractions lie in one of the intervals $]0, \frac{1}{2}]$ resp. $\left.] \frac{1}{2}, 1\right]$, where two of these are either directly consecutive in the listing or the first and the last fraction are among them. In any case, the positive difference of these two fractions is smaller than $\frac{1}{2}$ and the four indices involved are pairwise distinct.

Now we show that $C \geq \frac{1}{2}$ holds. For this we consider the example $1, 2, 2, 2, r$, where $r$ is meant to be a gigantic number. With these numbers one can form - ordered by size - the fractions $\frac{1}{r}, \frac{2}{r}, \frac{1}{2}, \frac{2}{2}, \frac{2}{1}, \frac{r}{2}, \frac{r}{1}$, where according to the problem statement $\frac{1}{r}$ and $\frac{2}{r}$ may not be chosen simultaneously. Therefore the smallest positive difference equals $\frac{1}{2}-\frac{2}{r}$, and this approaches the value $\frac{1}{2}$ from below arbitrarily closely as $r \rightarrow \infty$. $\square$