We will do the following algorithm to rearrange the children.
- Put the children who attend exactly one club at the start of the line (in any order).
- Put the children who attend exactly 2 clubs at the start of the line (in any order), and so on.
- Finally, put the children who attend exactly 30 clubs at the start of the line (in any order).

Since each club contains 40 children, then the number of pairs (club, child) is equal to
$$1200 = 30 \times 40 = n_{1} + 2 n_{2} + \cdots + 30 n_{30}.$$

Now back to the line, we will count from $1 \rightarrow 1200$ and for each number, we will point at the children in the following way:
- We point at the children from the top to the bottom of the line.
- If a child attends $k$ clubs, we point at him $k$ times (each time, we also count the numbers).

Finally, we create the new clubs $C_{1}, C_{2}, \ldots, C_{40}$ and add children to them using the rule: if we count a number $i \bmod 40$ while pointing at some child, then put that one into $C_{i}$ for any $i$.

We will count $\frac{1200}{40} = 30$ numbers which are $i \bmod 40$ for each $i$, so indeed, each club has exactly 30 children, and no one appears in the same club since each child is counted at most 30 times in the row. And it is easy to check that the numbers $n_{1}, n_{2}, \ldots, n_{30}$ remain unchanged.