Elegant proof of combinatorial statement: at least $l-k+1$ children who were in larger teams There are $n$ children, who play a game where in round $1$ they spilt up in $k$ teams (every team non-empty, pairwise disjoint). After a while it gets boring, so they form new teams to play round $2$; this time, they split up in $l$ teams (every team non-empty, pairwise disjoint), where $l>k$, and this beeing the only condition.
Prove that there are at least $l-k+1$ children, for which their team in the first round was of strictly larger size than in the second round.
This was a homework problem in a course last year which I had a very hard time proving, and finally, a day before the deadline, I found a proof which was three pages long and used a pretty messy induction with a whole lot of definitions and notations I had to introduce (it took me the whole day to write).
Is there an elegant proof of this statement?
If you want to see my proof, comment below and I will try to summarize it. But I guarantee you, its very messy! :)

Say a child in a team of size $m$ takes up $1/m$ of a team. Say child $i$ takes up $x_i$ of a team in round $1$ and $y_i$ of a team in round $2$. Now we need to show there are at least $l-k+1$ values of $i$ for which $y_i>x_i$.
But $\sum_ix_i=k$ and $\sum_iy_i=l$, so $\sum_i(y_i-x_i)=l-k$. Each term in this sum is strictly less than $1$, so there must be strictly more than $l-k$ positive terms. 
