(3) On each vertex of a regular $n$-sided polygon, there is a magpie. Due to a sudden fright, all the magpies fly away. After some time, they all return to these points, with one magpie on each vertex, but not necessarily to their original vertices. Find all positive integers $n$, such that there must exist 3 magpies, for which the triangles formed by their respective initial and final vertices are either all acute, all right, or all obtuse.
Therefore, we know that $n=5$ does not work.
When $n=2 m, m \geqslant 3, m \in \mathbf{N}$, take a pair of opposite vertices $A$ and $B$ of the regular $n$-sided polygon and examine the landing points of two magpies (still called $A$ and $B$) flying from these two points. If the landing points of magpies $A$ and $B$ are still a pair of opposite vertices, then we can choose another magpie $C$. Thus, the vertices occupied by magpies $A, B, C$ before and after flying form a right-angled triangle. If the landing points $A^{\prime}, B^{\prime}$ of magpies $A$ and $B$ are not opposite vertices, then we can denote the opposite vertex of $A^{\prime}$ as $C^{\prime}$, and the magpie that lands on $C^{\prime}$ after returning is from point $C$. Thus, the vertices occupied by magpies $A, B, C$ before and after flying form a right-angled triangle. This proves that all even numbers $n$ not less than 4 satisfy the requirement.

Let $n=2 m+1, m \geqslant 3, m \in \mathbf{N}$. In this case, there are no right-angled triangles in the graph. Let's prove the following lemma.

Lemma: When $n=2 m+1, m \geqslant 3$, in a regular $n$-sided polygon with all diagonals connected, the number of obtuse triangles is more than half of the total number of triangles.

Proof of the lemma: Let the regular $n$-sided polygon be $A_{1} A_{2} \cdots A_{n}$. At this time, the number of obtuse triangles with diagonal $A_{1} A_{j+1}$ as the longest side is $j-1$. Therefore, the total number of obtuse triangles is $S=(2 m+1)(1+2+\cdots+(m-1))=\frac{1}{2}(2 m+1) m(m-1)$.
The total number of triangles in the regular $n$-sided polygon is
$$
t=\mathrm{C}_{n}^{3}=\frac{1}{3}(2 m+1) m(2 m-1)
$$

Since
$$
\begin{aligned}
m-1>\frac{1}{3}(2 m-1) \Leftrightarrow & 3 m-3>2 m-1 \Leftrightarrow \\
& m>3-1=2
\end{aligned}
$$

The latter is clearly true, so the lemma holds.
By the lemma, when $n=2 m+1, m \geqslant 3$, there always exist 3 magpies such that the 3 vertices they occupy before and after flying form an obtuse triangle. This indicates that all odd numbers not less than 7 satisfy the requirement in the problem.
In summary, the natural numbers that satisfy the requirement in the problem are
$$
n \geqslant 3, n \neq 5
$$

Solution 2: The proof up to this point is the same as Solution 1, except for the case $n=2 m+1, m \geqslant 3$ which is proven as follows.

Since the shape of the triangle does not change when the regular $n$-sided polygon, along with the magpies on its vertices, is rotated, we can assume that a magpie $A$ returns to the same vertex after flying away. Draw a diameter $A A^{\prime}$ of the circumcircle of the regular $n$-sided polygon through the vertex where $A$ is located. Thus, there are $m$ magpies on each side of the diameter. It is easy to see that any two magpies on the same side of the diameter, together with $A$, form an obtuse triangle. Since $m \geqslant 3$, among the $m$ magpies on the same side of the diameter, there are always two, $B$ and $C$, that remain on the same side after flying away and returning. Thus, the vertices occupied by magpies $A, B, C$ before and after flying form an obtuse triangle. This proves that all odd numbers not less than 7 satisfy the requirement in the problem;
Therefore, the natural numbers that satisfy the requirement in the problem are $n \geqslant 3, n \neq 5$.