It suffices to show that we can swap the order of any two consecutive children. This could only change anything if one of them is a girl and the other is a boy. Suppose that before any of the two comes to the pile, there are $r$ children in the room and $k =ar+b$ sweets in the pile, where $0<b\leq r.$ If the boy comes first, he takes $a+1$ sweets and so $a(r-1)+b-1$ sweets remain, so the girl receives $a$ sweets (if $b-1 <r-1$ i.e. $b<r$) and $a+1$ otherwise. On the other hand, if she goes first, she takes $a$ sweets (if $b<r$) and $a+1$ otherwise. In the first case, there are $a(r-1)+b$ sweets left, so the boy gets again $a+1$ sweets since in this situation we have $b\leq r-1$. In the second case, there are $(a+1)(r-1)$ sweets so the boy again takes $a+1$ sweets. Hence changing the order does not change the final distribution and so we are done.