59th Ukrainian National Mathematical Olympiad

First of all, note that this value can be achieved with the following fractions choice: $$\frac{1}{1011},\frac{2}{1012},\frac{3}{1013},\dots ,\frac{1010}{2019}.$$

For the sake of contradiction, let's suppose that the smaller value of the maximum fraction can be obtained in some other way. Clearly, $2019$ cannot be a numerator, hence we can consider the fraction with denominator $2019$. Its numerator should be smaller than $1010$. Let's denote it as $a<1010$. There are at least $1009$ numbers in the set $$M=\{a,a+1,\dots ,1010,1011,\dots ,2018\}.$$ From the pigeonhole principle it follows that some of these numbers form one of the rest $1008$ (excluding $\frac{a}{2019}$) fractions. If we denote them $b<c$ then we will have $\frac{1010}{2019}<\frac{b}{2019}<\frac{b}{c}$, a contradiction.