Let $f(k)$ be the the $k$-th element in $S_i$ position in $S_{i+1}$. The operation can be written as \[ f(k)= \begin{cases} \frac{k+1}{2}, \text{ if } k \equiv 1 \pmod 2 , \\ \frac{k}{2}+n, \text{ if } k \equiv 0 \pmod 2 . \end{cases} \]Note that $a_1$ and $a_{2n}$ never change positions, we only need to care about $2 \le k \le 2n-1$, then the above relation can be combined as \[ f(k) \equiv \frac{k+1}{2} \pmod{2n-1} \Longrightarrow 2(f(k)-1) \equiv k-1 \pmod{2n-1}. \]Let $r=\varphi(2n-1) <2n$, then \[ f^{(r)}(k)-1 \equiv 2^r(f^{(r)}(k)-1) \equiv k-1 \pmod{2n-1}, \]i.e. $f^{(r)}(k)=k$. Therefore, $T \le r <2n$.