The desired value is $C=\frac{1}{2}$.
First, we prove that $C \leq \frac{1}{2}$. To do this, we assume without loss of generality (oBdA) 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, at least three of these fractions lie in one of the intervals ]0, $\frac{1}{2}$ ] or $\left.] \frac{1}{2}, 1\right]$. Among these, two fractions are consecutive in the list or the first and the last fraction are included. In any case, the positive difference between these two fractions is less than $\frac{1}{2}$, and the four involved indices are pairwise distinct.
Now we show that $C \geq \frac{1}{2}$. For this, consider the example 1, 2, 2, 2, $r$, where $r$ is a very large number. With these numbers, we can form the fractions $\frac{1}{r}, \frac{2}{r}, \frac{1}{2}, \frac{2}{2}, \frac{2}{1}, \frac{r}{2}, \frac{r}{1}$, ordered by size. According to the problem statement, $\frac{1}{r}$ and $\frac{2}{r}$ cannot both be chosen. Therefore, the smallest positive difference is $\frac{1}{2}-\frac{2}{r}$, which approaches the value $\frac{1}{2}$ from below as $r \rightarrow \infty$. $\square$