Sum of fractions till 2019

Note that $\frac{1}{3+6+\ldots+3r} = \frac{1}{3}\times\frac{1}{1+2+\ldots+r} = \frac{1}{3} \times \frac{2}{r(r+1)} = \frac{2}{3}\left(\frac{1}{r}-\frac{1}{r+1}\right)$ and $2019 = 3\times 673$ .

It means that the desired (telescoping) sum is $\frac{2}{3} \left(\frac{1}{1} -\frac{1}{674} \right) = \frac{673}{1011} = \frac{a}{b}$ . Hence $a+b=1684$ .

Similar solution with @Chan Lye Lee 's

$\begin{aligned} S & = \frac 13 + \frac 1{3+6} + \frac 1{3+6+9} + \cdots + \frac 1{3+6+9+\cdots + 2019} \\ & = \frac 13 \left(\frac 11 + \frac 1{1+2} + \frac 1{1+2+3} + \cdots + \frac 1{1+2+3+\cdots + 673}\right) \\ & = \frac 13 \sum_{n=1}^{673} \frac 1{\color{#3D99F6}\sum_{k=1}^n k} & \small \color{#3D99F6} \text{Sum of a GP} \\ & = \frac 13 \sum_{n=1}^{673} \frac 2{n(n+1)} \\ & = \frac 23 \sum_{n=1}^{673} \left(\frac 1n - \frac 1{n+1}\right) \\ & = \frac 23 \left(\frac 11 - \frac 1{674}\right) \\ & = \frac {673}{1011} \end{aligned}$

Therefore $a+b = 673+1011 = \boxed{1684}$ .