Solution:

Rearrange the children in the back row into order, and rearrange the front row in the same way, so that each child stays in front of the same child in the back row. Denote heights in the back row by $a_i$ and heights in the front row by $b_i$. So we have $a_1 \leq a_2 \leq \ldots \leq a_n$, and $a_i > b_i$ for $i = 1, 2, \ldots, n$.

Now if $i < j$, but $b_i > b_j$, then we may swap $b_i$ and $b_j$ and still have each child taller than the child in front of him. For $b_i < a_i \leq a_j$, and $b_j < b_i < a_i$. By repeated swaps we can get the front row into height order. [For example, identify the shortest child and swap him to the first position, then the next shortest and so on.]