The common divisors of $225$ and $105$ are $1$, $3$, $5$ and $15$. Each of these numbers can represent the number of groups that can be formed from the children in order for each group to get equal number of balls. So the problem has $4$ solutions:

- $1$ group with $225$ children and $105$ balls
- $3$ groups with $75$ children and $35$ balls each
- $5$ groups with $45$ children and $21$ balls each
- $15$ groups with $15$ children and $7$ balls each.