Solution:

We show that $T_{n}-n$ is even. Note that the subsets $\{1\},\{2\}, \cdots,\{n\}$ are good. Among the other good subsets, let $A$ be the collection of subsets with an integer average which belongs to the subset, and let $B$ be the collection of subsets with an integer average which is not a member of the subset. Then there is a bijection between $A$ and $B$, because removing the average takes a member of $A$ to a member of $B$; and including the average in a member of $B$ takes it to its inverse. So $T_{n}-n=|A|+|B|$ is even.

Solution:

Let $S=\{1,2, \ldots, n\}$. For a subset $A$ of $S$, let $\bar{A}=\{n+1-a \mid a \in A\}$. We call a subset $A$ symmetric if $\bar{A}=A$. Note that the arithmetic mean of a symmetric subset is $(n+1)/2$. Therefore, if $n$ is even, then there are no symmetric good subsets, while if $n$ is odd then every symmetric subset is good.

If $A$ is a proper good subset of $S$, then so is $\bar{A}$. Therefore, all the good subsets that are not symmetric can be paired. If $n$ is even then this proves that $t_{n}$ is even. If $n$ is odd, we have to show that there are odd number of symmetric subsets. For this, we note that a symmetric subset contains the element $(n+1)/2$ if and only if it has odd number of elements. Therefore, for any natural number $k$, the number of symmetric subsets of size $2k$ equals the number of symmetric subsets of size $2k+1$. The result now follows since there is exactly one symmetric subset with only one element.