Let the number of people be $ n$. Think of the guests as elements of $ \mathbb{Z}/n\mathbb{Z}$. For a permutation to satisfy the conditions of the problem and element $ i$ must be sent to $ i+x_i$ where the $ x_i$ are distinct integers from $ 1$ to $ n$. This permutation must be a union of disjoint cycles since it is finite. For $ (a_1,a_2,...,a_k)$ to be a cycle, we must have $ x_{a_1}+x_{a_2}+\cdots+x_{a_k}$ divisible by $ n$. But then we have the sum of all $ x_i$ divisible by $ n$. However, $ \sum x_i=1+2+\cdots+n=\frac{n(n-1)}{2}$ which is not divisible by $ n$ for even $ n$.

Another one: Let $f$ be this permutation of $S=\{1,...,2n\}$. Let $A=\{f(a)+a\pmod {2n}|a\in S\}$, $B=\{f(a)-a\pmod {2n}|a\in S\}$. I claim that one of $A,B$ is not $S$. Suppose $A=B=S$. Then $\exists a$ s.t. $f(n)+n\equiv f(a)-a\pmod {2n}$. But then $f(n)+n\equiv f(n)-n\equiv f(a)-a\pmod {2n}$, hence $|B|\le 2n-1 \Rightarrow B\ne S$, contradiction. Hence one of them is not $S$, so $\exists a,b\in S$ s.t. $f(a)\pm a\equiv f(b)\pm b\pmod {2n} \Rightarrow f(a)-f(b)\equiv \mp a\pm b\equiv \mp (a-b)\pmod {2n}$. It remains to note that this implies $|f(a)-f(b)|$ is either $a-b$ or $2n-(a-b)$, which is what we wanted.