Wow. Trigonometry or Calculus?

Use identity, $sin^2\theta = \dfrac {1-cos2\theta}{2}$ . All cosines add up to zero.

$\begin{aligned} \sum_{x = 1}^{2018} \left( \sin^2 \left(\cfrac{x \cdot \pi}{2018} \right) \right) = \sum_{x = 0}^{2018} \left( \sin^2 \left(\cfrac{x \cdot \pi}{2018} \right) \right) & \approx \lim_{\delta x \rightarrow 0} \sum_{x = 0}^{2018} \left( \sin^2 \left(\cfrac{x \cdot \pi}{2018} \right) \delta x \right) \\ & \approx \int_{0}^{2018} \left( \sin^2 \left(\cfrac{x \cdot \pi}{2018} \right) \right) \\ & \approx \boxed{1009} \end{aligned}$