2. Find the smallest real number $C$, such that for any positive real numbers $a_{1}, a_{2}, \cdots, a_{5}$ (allowing duplicates), we can always choose different indices $i, j, k, l$ such that $\left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C$.
2. The smallest real number $C=\frac{1}{2}$.

First, prove: $C \leqslant \frac{1}{2}$.
For positive real numbers $a_{1}, a_{2}, \cdots, a_{5}$, without loss of generality, assume $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$.
Consider the five ratios $\frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, \frac{a_{1}}{a_{5}}, \frac{a_{2}}{a_{3}}, \frac{a_{4}}{a_{5}}$, each of which lies in the interval $(0,1]$. By the pigeonhole principle, at least three of these ratios are in the interval $\left(0, \frac{1}{2}\right)$ or $\left(\frac{1}{2}, 1\right]$. In particular, there must be two adjacent ratios in an interval of length $\frac{1}{2}$ (considering $\frac{a_{1}}{a_{2}}$ and $\frac{a_{4}}{a_{5}}$ as adjacent). Therefore, the absolute difference between these two ratios is less than $\frac{1}{2}$.

Since the indices in these two adjacent ratios are all different, we can choose four indices $i, j, k, l$.
Thus, $C \leqslant \frac{1}{2}$.
Next, prove: The smallest real number $C=\frac{1}{2}$.
Consider the real numbers $1, 2, 2, 2, n$, where $n$ is a sufficiently large real number. The ratios of any two of these five numbers, arranged in increasing order, are
$$
\frac{1}{n}, \frac{2}{n}, \frac{1}{2}, \frac{2}{2}, \frac{2}{1}, \frac{n}{2}, \frac{n}{1}.
$$

Since $i, j, k, l$ are all different, $\frac{1}{n}$ and $\frac{2}{n}$ cannot both be selected. Therefore, the minimum absolute difference between any two ratios is $\frac{1}{2}-\frac{2}{n}$. Since $n$ can approach infinity, $\frac{1}{2}-\frac{2}{n}$ approaches $\frac{1}{2}$.
In conclusion, the smallest real number $C=\frac{1}{2}$.