28.2.11 ** Let $M$ be a set of $n$ points in the plane, satisfying:
(1) $M$ contains 7 points which are the 7 vertices of a convex heptagon;
(2) For any 5 points in $M$, if these 5 points are the 5 vertices of a convex pentagon, then this convex pentagon contains at least one point from $M$ inside it.
Find the minimum value of $n$.
Parse the proof that $n \geqslant 11$.
Consider a convex heptagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ with vertices in $M$. Connect $A_{1} A_{5}$. By condition (1), in the convex pentagon $A_{1} A_{2} A_{3} A_{4} A_{5}$, there is at least one point in $M$, denoted as $P_{1}$. Connect $P_{1} A_{1}$ and $P_{1} A_{5}$. Then, in the convex pentagon $A_{1} P_{1} A_{5} A_{6} A_{7}$, there is at least one point in $M$, denoted as $P_{2}$, and $P_{2}$ is different from $P_{1}$. Connect the line $P_{1} P_{2}$. Among $A_{1}, A_{2} \cdots, A_{7}$, at least 5 vertices are not on the line $P_{1} P_{2}$. By the pigeonhole principle, on one side of the line $P_{1} P_{2}$, there must be 3 vertices. These 3 vertices, together with points $P_{1}$ and $P_{2}$, form a convex pentagon that contains at least one point in $M$, denoted as $P_{3}$.

Next, draw the lines $P_{1} P_{3}$ and $P_{2} P_{3}$. Let the region $\pi_{3}$ be the half-plane bounded by the line $P_{1} P_{2}$ and on the side opposite to $\triangle P_{1} P_{2} P_{3}$ (excluding the line $P_{1} P_{2}$). Similarly, define regions $\pi_{1}$ and $\pi_{2}$. Thus, the three regions $\pi_{1}$, $\pi_{2}$, and $\pi_{3}$ cover all points on the plane except for $\triangle P_{1} P_{2} P_{3}$. By the pigeonhole principle, among the 7 vertices $A_{1}, A_{2}, \cdots, A_{7}$, there must be $\left[\frac{7}{3}\right]+1=3$ vertices in the same region (say $\pi_{3}$). These 3 points, together with $P_{1}$ and $P_{2}$, form a convex pentagon with vertices in $M$, and its interior contains at least one point in $M$, denoted as $P_{4}$. Therefore, $n \geqslant 11$.

Next, we construct an example to show that $n=11$ is possible. Consider a convex heptagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ as a lattice heptagon, with 4 points inside it. Condition (1) is clearly satisfied. This point set $M$ also satisfies condition (2). Proof by contradiction. Assume there exists a lattice convex pentagon whose interior does not contain any lattice points. Since the area of any lattice polygon can be expressed as $\frac{n}{2}\left(n \in \mathbf{N}^{*}\right)$, by the principle of the smallest number, there must be a lattice convex pentagon $A B C D E$ with the smallest area that does not contain any lattice points in its interior. This is because, considering the parity of the coordinates of the vertices, there are only 4 cases: (odd, even), (even, odd), (odd, odd), (even, even). Thus, among the vertices of the pentagon $A B C D E$, there must be two vertices with the same parity in both coordinates, and the midpoint $P$ of the line segment connecting these two vertices is still a lattice point. Since $P$ is not in the interior of the convex pentagon $A B C D E$, $P$ must be on one of the sides of the pentagon. Without loss of generality, assume $P$ is on side $A B$, then $P$ is the midpoint of $A B$. Connect $P E$, then $P B C D E$ is a lattice convex pentagon with a smaller area that does not contain any lattice points in its interior, which is a contradiction.
In conclusion, the minimum value of $n$ is 11.