3. Find all positive integers $n$ such that the set $\{1,2, \cdots, n\}$ has $n$ pairwise distinct three-element subsets $A_{1}, A_{2}, \cdots, A_{n}$, satisfying that for any $k(1 \leqslant i<j \leqslant n)$, we have $\left|A_{i} \cap A_{j}\right| \neq 1$.
(Feng Zhigang)
3. (1) When $n=4 k\left(k \in \mathbf{N}_{+}\right)$, construct $A_{1}$, $A_{2}, \cdots, A_{4 k}$ as follows:
For any $1 \leqslant i \leqslant k, 0 \leqslant j \leqslant 3$,
$A_{4 i-j}=\{4 i-3,4 i-2,4 i-1,4 i\} \backslash\{4 i-j\}$.
(2) When $n \neq 4 k\left(k \in \mathbf{N}_{+}\right)$, assume the set $\{1,2, \cdots, n\}$ has $n$ distinct three-element subsets $A_{1}$, $A_{2}, \cdots, A_{n}$, satisfying the conditions of the problem.

For $A_{1}=\{a, b, c\}$, consider all subsets that have a non-empty intersection with $A_{1}$, say $A_{2}, A_{3}, \cdots, A_{m}$, and denote
$U=A_{1} \cup A_{2} \cup \cdots \cup A_{m}$.
If $|U|=3$, then $m=1m$.
According to the above analysis, the sets $A_{1}, A_{2}, \cdots, A_{n}$ can be divided into several groups, where sets within the same group have non-empty intersections, and sets from different groups have empty intersections. In each group, the number of sets is no greater than the number of elements used. Since the total number of sets equals the number of elements in $\{1,2, \cdots, n\}$, each group must have exactly 4 sets, which contradicts the fact that $n$ is not a multiple of 4.
In conclusion, the required positive integer $n$ is all multiples of 4.