Answer: D

Solution:

On each interval where $\sin nx\ge 0,$ the graphs of $\sin x$ and $\sin nx$ meet twice, unless they share the value $1$ there, in which case they meet once. Counting the humps and the endpoint at $(\pi,0)$ gives

$F(n)=n+1$ when $n$ is even or $n\equiv 3\pmod 4,$ and $F(n)=n$ when $n\equiv 1\pmod 4.$

Thus $$\begin{aligned} &\sum_{n=2}^{2007}F(n) \\ &=\sum_{n=2}^{2007}(n+1) \\ &\quad {}-\#\{n\equiv 1\!\!\pmod 4\}. \end{aligned}$$ The first sum is $2{,}017{,}033,$ and there are $501$ values $n\equiv 1\pmod 4$ in the range, giving $2{,}017{,}033-501=2{,}016{,}532.$

Thus, the correct answer is D.