Let

Then
$$
\begin{array}{l}
y_{1}=a_{1}-a_{2}+a_{3}-a_{4}+\cdots-a_{n-1}+a_{n}, \\
y_{2}=a_{2}-a_{3}+a_{4}-a_{5}+\cdots-a_{n}+a_{1}, \\
\cdots \cdots \\
y_{n}=a_{n}-a_{1}+a_{2}-a_{3}+\cdots-a_{n-2}+a_{n-1} .
\end{array}
$$
$$
\begin{array}{l}
y_{1}+y_{2}=2 a_{1}, \\
y_{2}+y_{3}=2 a_{2},
\end{array}
$$
$\qquad$
$$
y_{i}+y_{i+1}=2 a_{i}
$$
$\qquad$
$$
y_{n}+y_{1}=2 a_{n} .
$$

For every $1 \leqslant i \leqslant n$,
$$
y_{i}=S-2\left(a_{i+1}+a_{i+3}+\cdots+a_{i+n-2}\right) \text {. }
$$

where $S=a_{1}+a_{2}+\cdots+a_{n}=\frac{1}{2} n(n+1), \quad a_{n+k}=a_{k}$.
(1) When $n=4 k-1, \quad k \geqslant 1$.
Since $S$ is even in this case, each $y_{i}$ is even.
There exists $1 \leqslant j \leqslant 4 k-1$, such that
$$
y_{j}+y_{j+1}=2 \quad\left(y_{4 k}=y_{1}\right),
$$

Thus, $y_{j}$ and $y_{j+1}$ cannot both be positive.
Therefore, $n=4 k-1$ is not a "good" odd number.
(2) When $n=4 k+1, \quad k \in N$.
Since $S$ is odd in this case, each $y_{i}$ is odd.
Let
$$
y_{1}=1, \quad y_{2}=3, \quad \cdots, \quad y_{2 k+1}=4 k+1,
$$
$$
\begin{aligned}
y_{2 k+2} & =4 k+1, \quad y_{2 k+3}=4 k-3, \quad y_{2 k+4}=4 k-3, \cdots, \quad y_{4 k-1}=5, \\
y_{4 k} & =5, \quad y_{4 k+1}=1 .
\end{aligned}
$$

Then $a_{1}=2, a_{2}=4, \cdots, a_{2 k}=4 k, a_{2 k+1}=4 k+1$,
$$
\begin{array}{l}
a_{2 k+2}=4 k-1, \quad a_{2 k+3}=4 k-3, \cdots, a_{4 k-1}=5, a_{4 k}=3, a_{4 k+1} \\
=1 .
\end{array}
$$

Thus, all $n=4 k+1, k \in N$ are "good" odd numbers.
$7 \cdot 70$ An odd number $n \geqslant 3$ is called a "good" odd number if and only if there exists a permutation $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ of $\{1,2, \cdots, n\}$ such that the following $n$ sums
$$
\begin{array}{l}
a_{1}-a_{2}+a_{3}-a_{4}+\cdots-a_{n-1}+a_{n}, \\
a_{2}-a_{3}+a_{4}-a_{5}+\cdots-a_{n}+a_{1}, \\
\cdots \cdots \\
a_{n}-a_{1}+a_{2}-a_{3}+\cdots-a_{n-2}+a_{n-1} .
\end{array}
$$

are all positive. Find all "good" odd numbers.
(32nd International Mathematical Olympiad Preliminary Problem, 1991)