Consider first the endpoints, A and K: The sum of the distances to these two points will be the same for any point between them. Similarly, any point between B and J will have the same total distance to points A, K, B, and J. Proceeding in this way and taking the points in pairs, we find that every point between E and G produces the same total distance to the outermost 10 labeled points. That leaves only F. So the answer is that point F has the minimum total distance to the labeled points.

Strangely, this means that we can shift the points around without affecting this result. If points A through E crowded F closely on the left, and point K were 90 light-years away on the right, point F would still yield the smallest total sum, so long as the points appeared in the order shown above.