Lots of 2019

Let the roots of the equation be $a_k$ , where $k=1, 2, 3, \dots 2019$ . Then $a_k^{2019} + 2019a_k + 2019 = 0$ . Now we have:

$\begin{aligned} \sum_{k=1}^{2019}a_k^{2019} & = \sum_{k=1}^{2019} \left(-2019a_k - 2019\right) \\ & = - 2019 {\color{#3D99F6}\sum_{k=1}^{2019} a_k} - \sum_{k=1}^{2019} 2019 & \small \color{#3D99F6} \text{By Vieta's formula} \\ & = - 2019 \times {\color{#3D99F6}0} - 2019^2 & \small \color{#3D99F6} \text{Since }2019 = 3\times 673, \\ & = - 3^2 \times 673^2 & \small \color{#3D99F6} \text{where 3 and 673 are primes} \end{aligned}$

Therefore, $a+b+c+d = 3+2+673+2 = \boxed{680}$ .

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Reference: Vieta's formula