= Solution
Let $e_k$ be the expected number of future children receiving stickers from a box containing $k$, with $e_0=0$. Conditioning on the first uniform remainder gives
$$
e_k=1+\frac1k\sum_{j=0}^{k-1}e_j.
$$
Subtract the corresponding identity for $k-1$ after multiplying each by its denominator. This gives $ke_k=k e_{k-1}+1$, so $e_k=e_{k-1}+1/k$. Therefore $e_N=\sum_{j=1}^N1/j$, and
$$
\boxed{e_4=1+\frac12+\frac13+\frac14=\frac{25}{12}.}
$$
The recurrence counts the current positive recipient once and excludes all later children arriving after the box is empty.
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