MOSP $2002$ Combinatorics Problem I only want a hint(I already have the solution near me, but the book doesn't give a hint

(MOSP) Assume that each of the $30$ MOPpers has exactly one favorite chess variant and exactly one favourite classical inequality. Each MOPper lists this information on a survey. Among the survey responses, there are exactly $20$ different chess variants and $10$ different classical inequalities. Let $n$ be the number of MOPpers $M$ such that the number of MOPpers who listed $M$'s favourite inequality is greater than the number of MOPpers who listed $M$'s favourite chess variant. Prove that $n\ge 11$. Source-$102$ Combinatorial Problems

The problem here is that I don't know how to attack this type of problem. The two things that this question reminds me of are :
1)Pigeonhole principle
2)Graph theoretic interpretation
The first one tells us that there are at least $3$ people liking the same inequality, and at least $2$ people liking the same chess variant. Graph theoretic interpretation seems to make things worse here. 
A hint will be appreciated (more than a complete solution anyway, since I already have it). Mentioning your thought process and motivation will be very much appreciated.
Thanks in advance.
P.S: Can anyone suggest a better heading? I can't find one.
 A: Number the MOPpers $1$ through $30$. For MOPper $k$ let $a_k$ be the number of MOPpers with the same favorite chess variant as MOPper $k$ (including MOPper $k$), and let $b_k$ be the number of MOPpers with the same favorite classical inequality as MOPper $k$ (again including MOPper $k$ in the count). We want to show that $a_k<b_k$ for at least $11$ of the $30$ MOPpers.
It’s tempting to consider $\sum_{k=1}^{30}a_k$ and $\sum_{k=1}^{30}b_k$: if we could get a good handle on them, we might be able to use
$$\sum_{k=1}^{30}b_k-\sum_{k=1}^{30}a_k=\sum_{k=1}^{30}(b_k-a_k)$$
to show that at least $11$ terms $b_k-a_k$ have to be positive.
It isn’t immediately obvious what these sums are, but if we make a bipartite graph $G$ with vertex sets $V$ for the $20$ chess variants and $W$ for the $10$ classical identities and $30$ edges, one for each MOPper, between the MOPper’s favorite chess variant and favorite classical identity, then 
$$\sum_{k=1}^{30}a_k=\sum_{v\in V}(\deg v)^2\qquad\text{and}\qquad\sum_{k=1}^{30}b_k=\sum_{v\in W}(\deg v)^2\;.$$
If $v\in V$, for instance, there are $\deg v$ MOPpers whose favorite chess variant is $v$, and if MOPper $k$ is one of them, then $a_k=\deg v$, so they contribute $(\deg v)^2$ to the first sum. 
Unfortunately, while we know that $\sum_{v\in V}\deg v=\sum_{v\in W}\deg v=30$, it’s hard to get a handle on the sums of the squares of the degrees. However, if we replace $a_k$ and $b_k$ by $\frac1{a_k}$ and $\frac1{b_k}$, respectively, we get something far more manageable: for $v\in V$ we still have $\deg v$ MOPpers whose favorite chess variant is $v$, but each of them contributes $\frac1{\deg v}$ to $\sum_{k=1}^{30}\frac1{a_k}$, so as a group they contribute $1$. (This is where we use the fact that all of the degrees are positive; there are counterexamples without that hypothesis.) Thus,
$$\sum_{k=1}^{30}\frac1{a_k}=20\qquad\text{and}\qquad\sum_{k=1}^{30}\frac1{b_k}=10\;,$$
and
$$\sum_{k=1}^{30}\left(\frac1{a_k}-\frac1{b_k}\right)=10\;.\tag{1}$$
Moreover, $b_k>a_k$ iff $\dfrac1{a_k}-\dfrac1{b_k}>0$, so the problem reduces to showing that at least $11$ terms of $(1)$ must be positive.
I’ll stop here: replacing $a_k$ and $b_k$ by their reciprocals was the key insight, and while the rest may take some thought, it’s comparatively straightforward.
