3. For any sequence, the following operation can be performed: each time select three consecutive terms, denoted as $a, b, c$, and replace them with $b, c, a$, keeping the other terms unchanged. Determine all integers $n \geqslant 3$, such that the sequence $1, 2, \cdots, n$ can be transformed into $n, n-1, \cdots, 1$ after a finite number of operations.
（Liang Yingde, problem contributor）
3. For an integer $n \geqslant 3$, the condition is satisfied if and only if
$n \equiv 0$ or $1(\bmod 4)$.
On one hand, for a sequence $\sigma_{1}, \sigma_{2}, \cdots, \sigma_{m}$ consisting of $m$ different numbers, define the "inversion number" of the sequence:
$I\left(\sigma_{1}, \sigma_{2}, \cdots, \sigma_{m}\right)$
$=\mid\left\{(j, k) \mid 1 \leqslant j\sigma_{k}\right\}$.
If $\sigma$ is transformed into $\tau$ by one operation as described in the problem, it is easy to verify that
$$
I(\tau)-I(\sigma)=I(b c a)-I(a b c) \in\{-2,0,2\} .
$$

In particular, the parity of $I(\sigma)$ remains unchanged after one operation. Therefore, if it is possible to transform $1,2, \cdots, n$ into $n, \cdots, 2,1$ through several operations, then $I(1,2, \cdots, n)=0$ and $I(n, \cdots, 2,1)=\frac{n(n-1)}{2}$ must have the same parity, i.e.,
$n \equiv 0$ or $1(\bmod 4)$.
On the other hand, prove that:
When $n \equiv 0$ or $1(\bmod 4)$, it is possible to transform $1,2, \cdots, n$ into $n, \cdots, 2,1$ through several operations.
Use induction on $n$ to prove the conclusion (1).
When $n=4$, the following operations can be used:
$1234 \rightarrow 2314 \rightarrow 2143 \rightarrow 2431 \rightarrow 4321$.
Assume $n4$.
Notice that one operation $a b c \rightarrow b c a$ moves $a$ from the left side of $b c$ to the right side. First, perform this operation twice to move $m-4$ to the right of $m-3, m-2, m-1, m$, then perform this operation twice to move $m-5$ to the right of $m-3, m-2, m-1, m$, and so on, until moving 1 to the right of $m-3, m-2, m-1, m$, resulting in the sequence $\square$
$$
m-3, m-2, m-1, m, 1,2, \cdots, m-4 .
$$

Using the fact that the conclusion (1) holds for $n=4$, the first 4 terms $m-3, m-2, m-1, m$ can be transformed into reverse order through several operations. For the remaining $m-4$ terms, note that
$m-4 \equiv 0$ or $1(\bmod 4)$.
If $m-4=1$, no further operations are needed; if $m-4 \geqslant 4$, by the induction hypothesis, $1,2, \cdots, m-4$ can be transformed into reverse order through several operations.
In summary, the conclusion (1) holds for $n=m$.