Solution:

Each box must either contain a single group of four consecutive balls (e.g. $5$, $6$, $7$, $8$) or two groups of two consecutive balls (e.g. $5$, $6$, $9$, $10$). Since all groups have even lengths, this means that $1$ and $2$ are in the same group, $3$ and $4$ are in the same group, and so on. We can think of each of these $8$ pairs of balls as an individual unit, so the answer is equal to the number of ways to put $8$ objects in $4$ indistinguishable boxes, where each box has $2$ objects without any additional restrictions. The number of ways to do this is
$$\frac{8!}{2^{4}\cdot 4!} = \boxed{105}$$