Answer: 53

Small Hint:

$\frac{1 + \sqrt{3}i}{2}$ is a primitive sixth root of unity, and $2016$ is a multiple of $6$

Big Hint:

Matching imaginary parts gives $a + b = 4038;$ the bound on the coefficients then forces both values

Solution:

Let $\omega = \frac{1 + \sqrt{3}i}{2} = \cos 60^\circ + i\sin 60^\circ,$ a primitive sixth root of unity. Since $2016 = 6 \cdot 336,$ we get $\omega^{2016} = 1,$ $\omega^{2017} = \omega,$ and $\omega^{2018} = \omega^2 = \frac{-1 + \sqrt{3}i}{2}.$ Therefore $$\begin{aligned} f(\omega) &= a\omega^2 + b\omega + c \\ &= \left(c + \frac{b - a}{2}\right) \\ &\quad {}+ \frac{(a + b)\sqrt{3}}{2}\,i. \end{aligned}$$

Matching imaginary parts, $\frac{a + b}{2} = 2019,$ so $a + b = 4038.$ Since $a \le 2019$ and $b \le 2019,$ this forces $a = b = 2019.$ Matching real parts then gives $c + 0 = 2015,$ so $c = 2015.$

Hence $f(1) = a + b + c$ $= 4038 + 2015 = 6053,$ whose remainder upon division by $1000$ is $53.$