The answer is $2\lceil n/2\rceil$. We can always achieve this bound by the following method given a plane on which none of the rays lie, consider the rays which point upward or downward with respect to this plane. By pigeonhole, at least $\lceil n/2\rceil$ rays point (wlog) upward. Now, since they all point upward, there exists a plane which intersects them all. Consider a sphere which almost degenerates into this plane, (in other words with a big enough radius), so that its part which tends to infinity is upwards. Therefore, all of the rays will be intersected twice, for a total of $2\lceil n/2\rceil$. To prove this is an upper bound, consider the following construction. Consider $\lceil n/2\rceil$ lines. On $\lfloor n/2\rfloor$ lines, place two rays (with opposite verses), and on the other line (there is one iff $n$ is odd) place a single ray. On a line there can be at most two intersection, so we can have at most a total of $2(\lfloor n/2\rfloor)+2(\lceil n/2\rceil -\lfloor n/2\rfloor)=2\lceil n/2\rceil$, as wanted.