7.24 Find the smallest natural number $n$ with the following property: when any 5 vertices of a regular $n$-gon $S$ are painted 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.
[Solution 1] For a regular 13-sided polygon, when the vertices $A_{1}, A_{2}, A_{4}, A_{6}, A_{7}$ are painted red, none of the 13 axes of symmetry satisfy the problem's requirements (see figure). For regular 12, 11, and 10-sided polygons, when the above 5 vertices are red, the conclusion of the problem also does not hold. When the number of sides is no more than 9, the conclusion is obviously not valid. Therefore, the smallest natural number $n \geqslant 14$.

Draw a regular 14-sided polygon and all its diagonals, then these segments can be divided into 14 groups, with all segments in each group being parallel and each having 1 axis of symmetry. Among these 14 axes of symmetry, 7 pass through a pair of opposite vertices of the regular 14-sided polygon, and the other 7 each bisect a pair of opposite sides of the regular 14-sided polygon. In the first 7 groups of parallel lines, each group has 6 lines, and there are two vertices on the axis of symmetry, which can be considered degenerate parallel line segments, called the odd points of the group of parallel lines. In the latter 7 groups of parallel lines, each group has 7 line segments, with no odd points. There can be 10 lines connecting the 5 red vertices, called red line segments. Thus, the problem reduces to whether a group of parallel lines can be found that contains neither red lines nor red odd points. Considering red odd points and red lines as red elements, the problem is to prove that these 15 red elements can fall into at most 13 groups of parallel lines.

In a regular 14-sided polygon, all sides and diagonals have only 7 different lengths. Therefore, by the pigeonhole principle, at least two red lines have the same length.
(1) Suppose there are 3 red lines of the same length, then among these 3 lines, there are always two without a common endpoint, thus the quadrilateral formed by their 4 endpoints is an isosceles trapezoid or a rectangle. If it is a rectangle, then the two pairs of opposite sides each belong to a group of parallel lines, so there is a group of parallel lines without red elements. If it is an isosceles trapezoid, then the upper and lower bases of the red lines belong to the same group of parallel lines. In addition, 3 equal-length line segments have 6 endpoints, and there are only 5 red points, so there must be two equal-length red lines with one common endpoint. Clearly, this common endpoint is precisely the odd point of the group of parallel lines formed by the line connecting the other two endpoints. This again leads to a group of parallel lines without red elements.
(2) Suppose among the 10 red lines, the number of equal-length line segments is at most two, then there are at least 3 pairs of red lines of equal length.
(a) Suppose there is a pair of equal-length red lines without a common endpoint, then the 4 endpoints of the two lines form a rectangle or an isosceles trapezoid. Both cases have been discussed in (1). Here, it only needs to be pointed out that when it is an isosceles trapezoid, the two legs and the two diagonals each form a group of equal-length line segments. Since there are at least 3 pairs of equal-length red lines, there is another group of equal-length red lines, whose 4 endpoints are different from the 4 vertices of the trapezoid. This can also lead to two red elements belonging to the same group of parallel lines.
(b) Any pair of equal-length red lines have a common endpoint, then each point can have at most one pair of equal-length red lines, otherwise, a pair of equal-length red lines without a common endpoint can be found. Each pair of equal-length line segments with a common endpoint leads to two red elements belonging to the same group of parallel lines.

In summary, in any case, there is a group of parallel lines without red elements. Therefore, the smallest natural number $n=14$.
[Solution 2] The proof that for $n \leqslant 13$ the regular $n$-sided polygons do not meet the problem's requirements is the same as in Solution 1. Below, we prove that the regular 14-sided polygon has the property required by the problem.

In the regular 14-sided polygon $A_{1} A_{2} \cdots A_{14}$, there are 7 axes of symmetry that do not pass through any vertex but each bisect a pair of opposite sides. We classify $A_{i}$ as odd or even vertices based on the parity of the index $i$. Clearly, under the reflection with respect to any of these 7 axes of symmetry, each pair of symmetric vertices has different parity.

Suppose among the 5 red vertices, there are $m$ odd vertices, $0 \leqslant m \leqslant 5$, and thus $5-m$ even vertices. The number of lines connecting the red odd vertices to the red even vertices is
$$
m(5-m) \leqslant 6.
$$

Thus, the number of perpendicular bisectors of these lines is at most 6. Therefore, among the above 7 axes of symmetry, at least one does not perpendicularly bisect any of these 6 lines, and this axis is the one we seek. In conclusion, the smallest value of the number of sides $n$ is 14.