, 2018

The solution is based on a simple fact, that if you add the arithmetic mean of the sequence to the sequence, then the arithmetic mean of the sequence will not change, since
$$\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}=\frac{a_{1}+a_{2}+\cdots+a_{n}+\frac{a_{1}+a_{2}+\cdots+a_{n}}{n}}{n+1}$$
Denote by $\mu(A)$ the arithmetic mean of the elements of $A$. Denote
$$P=\left\{A \subset\{1,2, \ldots, n\} \quad: \quad \mu(A) \in \mathbb{Z}_{+}, \mu(A) \in A\right\}$$
and
$$Q=\left\{A \subset\{1,2, \ldots, n\} \quad: \quad \mu(A) \in \mathbb{Z}_{+}, \mu(A) \notin A\right\} .$$
We need to prove that $|P|+|Q|$ is even. For that we will prove $|P|=|Q|$. Really, take any set from $P$ and remove its arithmetic mean. We will get an element from $Q$. Take any set from $Q$ and add its arithmetic mean. We will get an element from $P$. Since arithmetic mean of the set is defined uniquely, so we have bijection between $P$ and $Q$.