Consider a largest such group $G$. Why another member cannot be adjoined to this group? 1. Because he received a hat from someone from $G$. But then the set $H$ of these people has $|H| \leq |G|$. 2. Because he sent his hat to someone in $G$. But then the set $L$ of these people makes a group with the required property, so it has $|L| \leq |G|$. Therefore $30 = |G \cup H \cup L| \leq |G| + |H| + |L| \leq 3|G|$, so $|G| \geq 10$.

Connect two people if at least one of them send hat to the other. Then we have at most 30 edges in this graph $G$. So we have at least ${30\choose 2}-30= 405$ edges in complement $G'$. Let say there is no group with 10 members such that no one in the group has received a hat from another one in the group. This is there is no $K_{10}$ in $G'$. By Turan theorem in $G'$ we can have at most ${(p-2) n^2\over 2(p-1)}$ where $p=10$ and $n=30$, that is ${8\cdot 900\over 18} = 400 $ edges. Contradiction!

We consider $10$ people group $G$.There are $\textstyle\binom {30}{10}$ group $G$.For any person $A\in G$,$f(A)$ denotes the number of hat which $A$ receive from other person in $G$.Let $S(G)=\sum_{A\in G}^{}f(A)$.We consider $G$ which minimize $S(G)$.If $S(G)=0$,we can prove.We suppose that $S(G)>0$.Then ∃$A\in G$,∃$B\in G$ s.t. $A$ receive a hat from $B$.Thus there are at most $9$ people other than $G$ who receive a hat from person in $G$.So there are at least $11$ people other than $G$ who don't receive a hat from person in $G$.If all $11$ people give a hat to person in $G$,there are $10$ people which $S=0$.This contradicts the minimality of $S(G)$.Hence there is a person $C$ in $11$ people who don't give a hat to person in $G$.We replace $B$ with $C$.Then $S$ strictly decreasing which is absurd.Therefore $S(G)=0$,and the proof is completed.$\blacksquare$