Let us split all fractions into two natural groups: those on the odd position and those on the even.
$$\begin{align*} a_1 &= \frac{1}{2019}, \\ a_2 &= \frac{1+2018+3}{2019+2+2017} = \frac{1+2021}{2019+2019}, \\ a_3 &= \frac{1+2018+3+2016+5}{2019+2+2017+4+2015} = \frac{1+2021+2}{2019+2019+2}, \dots, \\ a_{505} &= \frac{1+2018+3+2016+5+\dots+1012+1009}{2019+2+2017+4+2015+\dots+1008+1011} = \frac{1+2021+504}{2019+2019+504}. \\ b_1 &= \frac{1+2018}{2019+2} = \frac{2019}{2021}, \\ b_2 &= \frac{1+2018+3+2016}{2019+2+2017+4} = \frac{2019+2}{2021+2} = \frac{2019}{2021}, \\ b_3 &= \frac{1+2018+3+2016+5+2014}{2019+2+2017+4+2015+6} = \frac{2019+3}{2021+3} = \frac{2019}{2021}, \dots, \\ b_{505} &= \frac{1+2018+3+2016+5+\dots+1012+1009+1010}{2019+2+2017+4+2015+\dots+1008+1011+1010} = \frac{2019+505}{2021+505} = \frac{2019}{2021}. \end{align*}$$

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 $a_{k+1}$:
$$a_{k+1} - a_k = \frac{1+2021k}{2019+2019k} - \frac{1+2021(k-1)}{2019+2019(k-1)} = \frac{(1+2021k)(2019+2021(k-1)) - (1+2021k)(2019+2021k)}{(2019+2019(k-1))(2019+2019k)} > 0 \Leftrightarrow$$

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.
$$$$