, 2024

We will show these numbers exist. For that we define sequences $a_1, a_2, \dots, a_{2024}$ and $b_1, b_2, \dots, b_{2024}$ and then consider numbers $c_i = 2^{a_i} \cdot 5^{b_i}$ for $i = 1, 2, \dots, 2024$.
We choose the sequences $a_i$ and $b_i$ in such a way that $a_i$ is increasing, $b_i$ is decreasing, and the differences $a_i - a_j$ and $b_j - b_i$ were all mutually distinct for all indices $i > j$. This will be enough because
$$\frac{c_i}{c_j} = \frac{2^{a_i} \cdot 5^{b_i}}{2^{a_j} \cdot 5^{b_j}} = \frac{2^{a_i - a_j}}{5^{b_j - b_i}},$$
this number has a decimal expansion of a length $b_j - b_i$, whereas analogously, $\frac{c_j}{c_i}$ has a length of $a_i - a_j$.

We now construct the needed sequences, starting with $a_i$. We will do it inductively. Take $a_1 = 1, a_2 = 2$. When we have the numbers $a_1, a_2, \dots, a_i$, then by choosing $a_{i+1} = 2a_i$ we will achieve $a_{i+1} - a_i > a_i - a_1$, therefore all newly added differences will be higher than the previous ones.

We can construct $b_i$ similarly, starting at the end by taking $b_{2024} = a_{2024}$, then $b_{2023} = 2b_{2024}$, and so on. Since $b_{2023} - b_{2024} = a_{2024}$, all the differences in $b_i$ will be at least $b_{2023} - b_{2024} = a_{2024}$.

Remark: In our construction, $a_i = 2^{i-1}$ and $b_i = 2^{4049-i}$.