First let's assume the two boxes are different. The balls labelled $1$ and $2$ can be put into either box. All balls with even labels must then be put into the same box that the ball labelled $2$ goes. Thus any ball with label which has a common factor greater than $1$ with any one of those even labels must be put into the same box too. Therefore, we only need to consider those balls with labels greater than $50$. That leaves with $10$ balls labelled by the primes: $53$, $59$, $61$, $67$, $71$, $73$, $79$, $83$, $89$, $97$. Thus we have twelve entities, namely the ten prime numbers, the number $1$ and all the balls that must go with $2$. Each can be put into any of the two boxes. Taking away the $2$ cases where one of the boxes is empty, the number of ways is $2^{12} - 2 = 4094$. Since the two boxes are identical, we need to divide this number by $2$. Therefore the answer is $2047$.