$a+b+ab$

Solving for b gives $b=\frac{2019 - a}{a + 1}$ , or $b=\frac{2019}{a+1} - \frac{a}{a+1}$

Because b is an integer, $2019 \equiv a \pmod{a+1}$

$a \equiv -1 \pmod{a+1}$ for any positive integer value of $a$ , so $2019 \equiv -1 \pmod{a+1}$ . Therefore, $(a+1)$ is a divisor of $2020$

The factors of $2020$ are { $1, 2, 4, 5, 10, 20, 101, 202, 404, 505, 1010, 2020$ }, so these are the possible values of $(a+1)$ . However, $1$ doesn't work because $a$ would be $0$ (not positive), and $2020$ doesn't work because $b$ would be $0$ . Subtracting one from each of the remaining numbers leaves { $1, 3, 4, 9, 19, 100, 201, 403, 504, 1009$ } as the possible values of $a$ .

$1+3+4+9+19+100+201+403+504+1009 = \boxed{2,253}$

$a+b+ab=2019\Rightarrow a+b(a+1)=2019\Rightarrow a+1+b(a+1)=2020\Rightarrow (a+1)(b+1)=2020$

Adding 1 to both sides, it is rewritten as $(a+1)(b+1)=2020$ . So $a+1$ must be a factor of $2020$ .

All factors of $2020=2^2×5×101$ are $1,2,4,5,10,20,101,202,404,505,1010,2020$ .

Since a and b must be positive, both factors must be greater than 1, and the factorization 1×2020 is discarded.

The sum of all possible a then is equals $1+3+4+9+19+100+ 201+403+504+1009=\boxed{2253}$ .

$\begin{aligned} b&=\frac{2020}{a+1}-1\\ a+1&\mid2020\\ a&\in\{1,3,4,9,19,100,201,403,504,1009\}\qquad(a\notin\{0,2019\}\because a,b\in\mathbb Z^+)\\ \sum a&=2253 \end{aligned}$