59th Ukrainian National Mathematical Olympiad

Let us split all fractions into two natural groups: those on the odd position and those on the even. $$\begin{array}{rl}{a}_1& =\frac{1}{2019},\\ a_2& =\frac{1+2018+3}{2019+2+2017}=\frac{1+2021}{2019+2019},\\ a_3& =\frac{1+2018+3+2016+5}{2019+2+2017+4+2015}=\frac{1+2021+2}{2019+2019+2},\dots ,\\ a_{505}& =\frac{1+2018+3+2016+5+\cdots +1012+1009}{2019+2+2017+4+2015+\cdots +1008+1011}=\frac{1+2021+504}{2019+2019+504}.\\ b_1& =\frac{1+2018}{2019+2}=\frac{2019}{2021},\\ b_2& =\frac{1+2018+3+2016}{2019+2+2017+4}=\frac{2019+2}{2021+2}=\frac{2019}{2021},\\ b_3& =\frac{1+2018+3+2016+5+2014}{2019+2+2017+4+2015+6}=\frac{2019+3}{2021+3}=\frac{2019}{2021},\dots ,\\ b_{505}& =\frac{1+2018+3+2016+5+\cdots +1012+1009+1010}{2019+2+2017+4+2015+\cdots +1008+1011+1010}=\frac{2019+505}{2021+505}=\frac{2019}{2021}.\end{array}$$

In this way, all fractions denoted by $b_i$, $i=1,505$ are of the same value, $\frac{2019}{2021}$. Let us now compare values of $a_k$ and $a_{k+1}$: $$a_{k+1}-a_k=\frac{1+2021k}{2019+2019k}-\frac{1+2021(k-1)}{2019+2019(k-1)}=\frac{(1+2021k)(2019+2021(k-1))-(1+2021k)(2019+2021k)}{(2019+2019(k-1))(2019+2019k)}>0\iff$$

2021 + 2021 \cdot (k-1) + 2021^2 k + 2021^2 k(k-1) - 2021 - 2021 \cdot k - 2021^2(k-1) - 2021^2 k(k-1) = 2021^2 - 2021 > 0 Therefore, the largest fraction would be $\frac{2019}{2021}$ or $a_{505} = \frac{1+2021\cdot504}{2019+2019\cdot504}$. Let us compare them. \frac{1+2021\cdot504}{2019+2019\cdot504} - \frac{2019}{2021} = \frac{(1+2021\cdot504) \cdot 2021 - (2019+2019\cdot504) \cdot 2019}{(2019+2019\cdot504) \cdot 2021} < 0 \Leftrightarrow \\ 2021 + 2021 \cdot 2021 \cdot 504 - 2019^2 - 2019^2 \cdot 504 = \\ = 2021 + (2021^2 - 2019^2) \cdot 504 - 2019^2 = 2021 + 8080 \cdot 504 - 2019^2 = -2020 < 0. $$