Bulgarian Spring Tournament

No! There exists a natural number $s$ such that $2^s\equiv 1(\mathrm{mod}n)$ (for example $s=\phi (n)$ from Euler's theorem or because the power series of $2$ modulo $n$ is periodic). We have $2^{s-1}\equiv \frac{n+1}{2}(\mathrm{mod}n)$ and so $x=\frac{n+1}{2}\in S_n$, but $2x=n+1>n$ is not in $S_n$. Also, if $t\le \frac{n-1}{2}$ is of $S_n$, then $2t<n$ is also. Therefore, the smallest natural number $t$ such that $t\in S_n$ and $2t\notin S_n$, is $\frac{n+1}{2}$. Since this number is different for different $n$, we get what we asked for. $\square$