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! :)
 A: Let there be $n$ children. Supposed there is at least 1 child that is in a small or equal size group in round 1 than in round 2. Let's examine what happens with this child removed.
Case 1: The removed child's round 1 and round 2 group remains non-empty. This results in $n-1$ children but does not change $k$ or $l$.
Case 2: The removed child's round 1 becomes empty. This results in $n-1$ children, $k-1$ groups in round 1 and $l$ groups in round 2. Then the question is to prove that there are at least $l - (k-1) + 1 > l - k + 1$ children for which their team in the first round was of strictly larger size than in the second round.
Case 3: The removed child's round 1 and round 2 become empty. This results in $n$, $k$, $l$ all reduced by 1 and does not change the question.
Case 4: The removed child's round 2 becomes empty. This is impossible, since that means the round 2 group has a smaller size.
With the above 4 cases, we can conclude that if we remove all the children that is in a smaller or equal size group in round 1 than in round 2 and ends up with $n'$ children, $k'$ groups in round 1 and $l'$ groups in round 2, the resulting quantity $l' - k' + 1 \ge l - k + 1$.
All there's left to do is to look at $n'$ children with all children in round 1 being in a larger group than in round 2. Let these children be in groups $K_1,...,K_{k'}$ in round 1 and $L_1,...,L_{l'}$ in round 2 (non-empty, pairwise disjoint). There are clearly at least $l' \ge l' - k' + 1 \ge l-k+1$ children.
A: 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. 
