3. Find the smallest natural number $n$ with the following property: when any five vertices of a regular $n$-gon $S$ are colored red, there is always a line of symmetry $l$ of $S$ such that the reflection of each red point across $l$ is not a red point. (Supplied by Hu Chengzhang, Nankai University)
When $n \leqslant 9$, a regular $n$-sided polygon clearly does not possess the property mentioned in the problem.

For a regular $n$-sided polygon $A_{1} A_{2} \cdots A_{n}$, when $n=2 k (k \in N)$, there are $2 k$ axes of symmetry: the lines $A_{i} A_{k+i} (i=1,2, \cdots, k)$ and the perpendicular bisectors of the segments $A_{i} A_{i+1} (i=1,2, \cdots, k)$. When $n=2 k+1 (k \in N)$, the lines connecting vertex $A_{i} (1 \leqslant i \leqslant 2 k+1)$ to the midpoint of segment $A_{k+i} A_{k+i+1}$ are the axes of symmetry.

When $n=10$, if the vertices $A_{1}, A_{2}, A_{4}, A_{6}$, and $A$ are colored red, the regular decagon does not possess the property mentioned in the problem. Let the lines connecting $A_{i} A_{5+i} (i=1,2, \cdots, 5)$ be denoted as $l_{i}$. Let the perpendicular bisectors of the segments $A_{i} A_{i+1} (i=1,2, \cdots, 5)$ be denoted as $l_{i+\frac{1}{2}} (i=1,2, \cdots, 5)$. Thus, the complete set of axes of symmetry for the regular decagon are $l_{1}, l_{2}, l_{3}, l_{1}, l_{5}, l_{\frac{3}{2}}, l_{\frac{5}{2}}, l_{\frac{7}{2}}, l_{\frac{9}{2}}, l_{\frac{1}{2}}$, a total of 10.

When $i=1,2,4$, $l_{i}$ maps point $A_{i}$ to itself. Therefore, $l_{1}, l_{2}, l_{4}$ are not the axes of symmetry required by the problem. $l_{3}$ maps point $A_{2}$ to point $A_{4}$, $l_{5}$ maps point $A_{4}$ to point $A_{6}$, $l_{\frac{3}{2}}$ maps point $A_{1}$ to point $A_{2}$, $l_{\frac{5}{2}}$ maps point $A_{1}$ to point $A_{4}$, $l_{\frac{7}{2}}$ maps point $A_{1}$ to point $A_{6}$, $l_{\frac{9}{2}}$ maps point $A_{2}$ to point $A_{7}$, and $l_{\frac{11}{2}}$ maps point $A_{4}$ to point $A_{7}$. Therefore, none of the 10 axes of symmetry satisfy the property mentioned in the problem. Thus, the regular decagon does not possess the property mentioned in the problem.

Similarly, it can be proven that when $n=11,12,13$, if the points $A_{1}, A_{2}, A_{4}, A_{6}, A_{7}$ are all colored red, these regular $n$-sided polygons do not possess the property mentioned in the problem.
Next, we prove that a regular 14-sided polygon possesses the property mentioned in the problem.
For a regular 14-sided polygon $A_{1} \wedge \mathrm{i} 2 \cdots A_{14}$, there are 7 axes of symmetry that do not pass through the vertices. When $i$ is odd, the point $A_{i}$ is called an odd vertex, and when $i$ is even, the point $A_{i}$ is called an even vertex. Clearly, each axis of symmetry that does not pass through a vertex makes odd vertices and even vertices symmetric to each other. Suppose there are $m$ red odd vertices $(m=0,1, \cdots, 5)$, then there are $5-m$ red even vertices. The number of lines connecting red odd vertices to red even vertices is
$$
m(5-m) \leqslant\left(\frac{5}{2}\right)^{2}<7.
$$

Since $m(5-m)$ is an integer, then $m(5-m) \leqslant 6$. This indicates that the number of perpendicular bisectors of segments connecting red odd vertices to red even vertices is at most 6. Therefore, there is at least one axis of symmetry that does not pass through a vertex and ensures that the symmetric point of any red point is not red.