First, prove 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 from $M$, 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 from $M$, denoted as $P_{2}$, and $P_{2}$ is different from $P_{1}$. Connect the line $P_{1} P_{2}$, then at least 5 of the vertices $A_{1}$, $A_{2}, \cdots, A_{7}$ are not on the line $P_{1} P_{2}$. By the pigeonhole principle, in the formed decagon, there is at least one point from $M$, 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 on the side of the triangle $P_{1} P_{2} P_{3}$ that does not contain the line $P_{1} P_{2}$ (including the line $P_{1} P_{2}$). Similarly, define the regions $\pi_{1}$ and $\pi_{2}$. These three regions $\pi_{1}$, $\pi_{2}$, and $\pi_{3}$ cover the plane except for the triangle $\triangle P_{1} P_{2} P_{3}$. By the pigeonhole principle, among the 7 vertices $A_{1}$, $A_{2}, \cdots, A_{7}$, at least $\left[\frac{7}{3}\right]+1=3$ vertices are in the same region (let's say $\pi_{3}$). These 3 points, along with $P_{1}$ and $P_{2}$, form a convex quadrilateral with vertices in $M$, and its interior contains at least one point from $M$, denoted as $P_{4}$. Therefore, $n \geqslant 11$.

Next, we construct an example to show that $n=11$ is possible. As shown in the figure, the heptagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ is a lattice heptagon with 4 points inside. 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 $\frac{n}{2} (n \in \mathbf{N})$, by the well-ordering principle, 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. Consider the parity of the coordinates of the vertices, there are 4 cases: (odd, even), (even, odd), (odd, odd), (even, even). Thus, the midpoints of the sides of the pentagon are all lattice points. Let $P$ be the midpoint of one of the sides, say $A B$. Then $P$ is the midpoint of $A B$, and 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.
3 Let $M$ be a set of $n$ points in the plane I, satisfying:
(1) $M$ contains 7 points that are the vertices of a heptagon;
(2) For any 5 points in $M$, these 5 points are the vertices of a convex pentagon, and this pentagon contains at least one point from $M$ inside it.
Find the minimum value of $n$. (Leng Gangsong)