A and B each select one number at random from the distinct numbers 1, 2, 3..., n and the probability that the number selected by A is less than the number selected by B is $\frac{1009}{2019}$. Now, the probability that the number selected by B is the number immediately next to the number selected by A is 

distinct  numbers 1, 2, 3, ....... n  A, B Selected one number each 

$\mathrm{n}\left(\mathrm{S}\right)=\mathrm{n}\times \mathrm{n}={\mathrm{n}}^2$

E=number Selected by A is less than the number selected by B 

$\begin{array}{rcl}\mathrm{n}\left(\mathrm{E}\right)& =& 1\times (\mathrm{n}-1)+1\times (\mathrm{n}-2)......+1\times 2+1\times 1\\ \mathrm{n}\left(\mathrm{E}\right)& =& \frac{(\mathrm{n}-1)n}{2 }\end{array}$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{E}\right)& =& \frac{\mathrm{n}-1}{2\mathrm{n}}=\frac{1009}{2019} \end{array}$

$\begin{array}{rcl}2019\mathrm{n}-2019 & =& 2018\mathrm{n}\\ & \Rightarrow & \mathrm{n}=2019\end{array}$

${\mathrm{E}}_1=$number Selected by B is number immediately next to the number selected by a

There  are 2018 consecutive  pairs up to 2019

$\mathrm{A}\left({\mathrm{E}}_1\right)=\frac{2018}{\left(2019\right) \left(2019\right)}$