28.2.13 Tai* rotates a regular $n$-sided polygon $M$ around its center by $\frac{\pi}{n}$, obtaining $M'$. How many convex polygons can the figure $M \cup M'$ be divided into at minimum?
Consider Figure $\mathrm{a}$, the shape $M \cup M^{\prime}$ can be divided into $n+1$ convex polygons. Below, we prove that it cannot be divided into fewer convex polygons. For this, we examine all the outer vertices (i.e., the vertices of the regular $n$-sided polygons $M$ and $M^{\prime}$) and number the convex polygons. Any outer vertex belongs to one or several convex polygons, and we color this outer vertex with the color corresponding to the smallest numbered polygon, say $i$. Thus, the problem is reduced to the following lemma:

Lemma: Suppose there are $m (\geqslant 2)$ points distributed on a circle, each point is colored with one of $k$ colors, and no two adjacent points have the same color. Furthermore, if points $A$ and $B$ are colored the same, and points $C$ and $D$ are colored the same but with a different color, then segments $AB$ and $CD$ do not intersect. Then the number of colors used $k_{m} \geqslant \left[\frac{m+3}{2}\right]$.

We prove the lemma by induction on $m$. When $m=2,3$, it is clear that $k=2,3$ respectively, and the claim holds. Assume the lemma holds for up to $m-1$ points, and consider the case of $m (\geqslant 4)$ points.

If these $m$ points are all different colors, then $k_{m}=m \geqslant \left[\frac{m+3}{2}\right]$. If there are two points of the same color, let $A$ and $B$ be such points, then $A$ and $B$ are not adjacent. As shown in Figure $\mathrm{b}$, let the neighbors of $A$ and $B$ in the counterclockwise direction on the circle be $X$ and $Y$ respectively. Draw a line $l$ that passes between $A$ and $X$, and also between $B$ and $Y$. Connect $A Y$ and $B X$, thus line $l$ divides the circle into two smaller "circles". One "circle" has $2 \leqslant m_{1} \leqslant m-2$ points, and the other has $2 \leqslant m-m_{1} \leqslant m-2$ points. The two same-colored points $A$ and $B$ are on different "circles", and by the problem's conditions and the choice of $l$, there cannot be two same-colored points on different "circles" (otherwise, their connecting line would intersect $AB$). Therefore, applying the induction hypothesis to the two "small circles" gives:
$$
k_{m} \geqslant k_{m_{1}} + k_{m-m_{1}} - 1 \geqslant \left[\frac{m_{1}+3}{2}\right] + \left[\frac{m-m_{1}+3}{2}\right] - 1 \geqslant \left[\frac{m+3}{2}\right].
$$

Thus, the lemma holds for all $m \geqslant 2$.

Since each color corresponds to one of the convex polygons, and $M \cup M^{\prime}$ has $2n$ outer vertices, the number of convex polygons is at least $\left[\frac{2n+3}{2}\right] = n+1$.