1. Find all positive integers $n$ $(n>1)$: If all positive integers less than $n$ and coprime with $n$ are denoted as $a_{1}<a_{2}<\cdots<a_{k}$, then $a_{i}+a_{i+1}(i=1,2, \cdots, k-1)$ are not divisible by 3.
1. $n=2,4$ or 10.

Direct verification shows that the only $n$ satisfying the conditions and not exceeding 28 are $2, 4, 10$.
Now assume $n>28$.
By $(m, n)=1$ if and only if $(n-m, n)=1$, we know $a_{i}+a_{k+1-i}=n(1 \leqslant i \leqslant k)$.
If $n$ is odd, then $a_{1}=1, a_{2}=2$.
Thus, 3 divides $\left(a_{1}+a_{2}\right)$.
If $n$ is even and $3 \mid n$, then take $i$ such that $a_{i}<\frac{n}{2}<a_{i+1}$.
Thus, 3 divides $a_{i}+a_{i+1}=n$. Therefore, $21 \mid n$ and $3 \nmid n$.
Thus, $a_{2}=3, a_{k-1}=n-3, a_{k}=n-1$.
Thus, if $n \equiv 2(\bmod 3)$, then $3\left|\left(a_{k-1}+a_{k}\right) \Rightarrow 3\right|(2 n-4)$.
Hence, $n \equiv 1(\bmod 3)$.
If there exists $i$ such that $a_{i}+a_{i+1} \equiv 2(\bmod 3)$, then $a_{k-i}+a_{k+1-i}=2 n-\left(a_{i}+a_{i+1}\right) \equiv 0(\bmod 3)$.
Thus, for $i=1,2, \cdots, k-1$, we have $a_{i}+a_{i+1} \equiv 1(\bmod 3)$.
Combining this with $a_{1}=1$, we get
$$
\left\{\begin{array}{l}
a_{1} \equiv a_{3} \equiv \cdots \equiv 1(\bmod 3), \\
a_{2} \equiv a_{4} \equiv \cdots \equiv 0(\bmod 3) .
\end{array}\right.
$$

Since $n$ is not coprime with 6 but is coprime with 9, we have $a_{4}=9$. Furthermore, by conclusion (1), we get $a_{3}=7$, i.e., $n$ is coprime with 7.
From the above, we know
$$
(n, 21)=(n, 27)=1 \text {, }
$$

But 22, 24, 26 are not coprime with $n$.
By conclusion (1), we know $5,23 \notin\left\{a_{1}, a_{2}, \cdots, a_{k}\right\}$, thus, 23, 25 are also not coprime with $n$.

Therefore, 21 and 27 are two consecutive terms in the sequence $a_{1}, a_{2}, \cdots, a_{k}$, but $3 \mid(21+27)$, which is a contradiction.
In summary, the only $n$ satisfying the conditions are $2,4,10$.