Solution:

Let $x = \tfrac{\pi y}{2}.$ Dividing by $2$ and using $\tfrac12(1 - \cos(2\pi y)) = \sin^2(\pi y),$ the equation simplifies to $$\cos(\pi y)\cos\left(\dfrac{4028\pi}{y}\right) = 1.$$

Both cosines must equal $1$ or both equal $-1,$ so $y$ and $\tfrac{4028}{y}$ are integers of the same parity. Since $4028 = 2^2 \cdot 19 \cdot 53$ is even, both must be even, so $y = 2a$ with $a$ a positive odd divisor of $2014 = 2 \cdot 19 \cdot 53,$ giving $a \in \{1, 19, 53, 19 \cdot 53\}.$

Each such $a$ gives $x = \tfrac{\pi y}{2} = \pi a,$ so the sum of solutions is $$\pi(1 + 19 + 53 + 19\cdot53) = \pi(19+1)(53+1) = 1080\pi.$$

Thus, the correct answer is D.