Let $A(n)$ - is the set of natural numbers $\{1,2, \dots ,n\}$.
Let $B$ - is any subset of $A(n)$. And $S(B)$ is the sum of all elements $B$.
Subset $B$ is "special subset" if $S(B)$ divisible by $2n$ ( Mod$[S(B),2n]=0$).
Example: $A(3)=\{ 1,2,3 \}$, so we have only two "special subset" - $\{\varnothing\}$ and $\{1,2,3\}$.
$A(5)=\{ 1,2,3,4,5 \}$, so we have $4$ "special subset" - $\{\varnothing\}, \{1,4,5\}, \{2,3,5\}, \{1,2,3,4\}$.
Let $F(n)$ is the number of all "special subsets" for $A(n)$, $n \in \mathbf{N}$.
I found for $n<50$ that $F(n)-1$ is the nearest integer to $\frac{2^{n-1}}{n}$. $F(n)$=Floor$[\frac{2^{n-1}}{n} + \frac{1}{2}] + 1$.
Is it possible to prove this formula for any natural $n$ -?