From the Arithmetic mean - Geometric mean Inequality,
$$\frac{1}{2019}(a_1 + a_2 + \dots + a_{2019}) \ge \sqrt[2019]{a_1 a_2 \dots a_{2019}}, \text{ or} \\ (a_1 + a_2 + \dots + a_{2019})^{2019} \ge 2019^{2019} a_1 a_2 \dots a_{2019}.$$
From problem statement,
$$(a_1 + a_2 + \dots + a_{2019})^{2019} = (\sqrt[2018]{2019^{2019}})^{2019} = (2019 \cdot \sqrt[2018]{2019})^{2019} = \\ = 2019^{2019} \cdot \sqrt[2018]{2019^{2019}} = 2019^{2019} a_1 a_2 \dots a_{2019},$$
Which means that in this Arithmetic mean - Geometric mean Inequality, equality holds, which is possible iff
$$a_1 = a_2 = \dots = a_{2019} = \frac{1}{\sqrt[2019]{2019 \cdot 2019^{2019}}} = \sqrt[2019]{2019^{2019}}.$$