Problem 1499

1. Define the operation $S_i$ as follows:
$$S_i = (a_1, a_2, \ldots, a_{2n}) \mapsto S_{i+1} = (a_1, a_3, a_5, \ldots, a_{2n-1}, a_2, a_4, a_6, \ldots, a_{2n})$$
This operation rearranges the sequence by taking all odd-indexed elements first, followed by all even-indexed elements.

2. Define the function $f(k)$ that maps the position $k$ in $S_i$ to its new position in $S_{i+1}$:
$$f(k) = \begin{cases} \frac{k+1}{2}, & \text{if } k \equiv 1 \pmod{2}, \\ \frac{k}{2} + n, & \text{if } k \equiv 0 \pmod{2}. \end{cases}$$

3. Note that $a_1$ and $a_{2n}$ never change positions. Therefore, we only need to consider the positions $2 \le k \le 2n-1$.

4. Combine the above relations into a single congruence:
$$f(k) \equiv \frac{k+1}{2} \pmod{2n-1}$$
This can be rewritten as:
$$2(f(k) - 1) \equiv k - 1 \pmod{2n-1}$$

5. Let $r = \varphi(2n-1)$, where $\varphi$ is Euler's totient function. Since $2n-1$ is odd, $\varphi(2n-1) < 2n$.

6. Consider the iterated function $f^{(r)}(k)$:
$$f^{(r)}(k) - 1 \equiv 2^r (f^{(r)}(k) - 1) \equiv k - 1 \pmod{2n-1}$$
This implies:
$$f^{(r)}(k) = k$$
Therefore, the sequence returns to its original order after $r$ operations.

7. Since $r = \varphi(2n-1) < 2n$, we have:
$$T \le r < 2n$$

The final answer is $\boxed{ T < 2n }$.