First, we prove that $n \ge 11$. Suppose a convex heptagon has its vertices in $M$ given by $A_1A_2A_3A_4A_5A_6A_7$. Using Condition (1), we get that there exists one point $P_1$ belonging to $M$ in the interior of convex pentagon $A_1A_2A_3A_4A_5$. Connecting $P_1A_1$ and $P_1A_5$, we obtain that there exists one point $P_2$ in convex pentagon $A_1P_1A_5A_6A_7$ so that $P_2$ belongs to $M$ and is different from $P_1$. Then, there are at least 5 points in $\{A_1, A_2, A_3, A_4, A_5, A_6, A_7\}$ which do not lie on line $P_1P_2$. By the Pigeon Hole Principle, there exist at least 3 points on one side of line $P_1P_2$, and these 3 points together with $P_1$ and $P_2$ constitute a convex pentagon which contains at least one point $P_3$ belonging to $M$.

Now, we have three lines $P_1P_2$, $P_2P_3$ and $P_3P_1$, which form a triangle $\triangle P_1P_2P_3$. Let $\pi_1$ denote the half-plane on one side of line $P_1P_2$ which is opposite to $\triangle P_1P_2P_3$ and contains no points on $P_1P_2$. In a similar way, we define $\pi_2$ and $\pi_3$. Areas $\pi_1$, $\pi_2$ and $\pi_3$ cover the entire plane except $\triangle P_1P_2P_3$. By the Pigeon Hole Principle, there is one area of $\pi_1$, $\pi_2$ and $\pi_3$ which contains at least 3 points belonging to $\{A_1, A_2, A_3, A_4, A_5, A_6, A_7\}$, Without loss of generality, we assume that the area $\pi_1$ contains points $A_1, A_2, A_3$, then there exists one point $P_4$ belonging to $M$ within the convex pentagon constituted by $A_1, A_2, A_3, P_1$ and $P_2$. So, $n \ge 11$.

Now, we give an example to illustrate that $n=11$ is attainable. As seen in the figure, set $M$ consists of integral points $A_1, A_2, A_3, A_4, A_5, A_6, A_7$, and four integral points within the heptagon $A_1A_2A_3A_4A_5A_6A_7$. Obviously, $M$ satisfies Condition (1). We are going to

prove that $M$ also satisfies Condition (2).
By reduction to absurdity, assume that there is a convex pentagon with its vertices belonging to $M$ which contains no point of $M$ in its interior. Then among such pentagons there must be one, denoted by $ABCDE$, which has the least area, since the value of the area of a polygon with integral vertices is always in the form of $\frac{n}{2}$ ($n \in \mathbb{N}$).
There are only 4 cases concerning the odd/even property of the $xy$-coordinate of an integral point: (odd, even), (even, odd), (odd, odd), (even, even). So there must be two vertices among $A, B, C, D, E$ which have the same odd/even property, and the midpoint of the segment formed by these two vertices, say $P$, is also an integral point and belongs to $M$. By definition, $P$ is not in the interior of pentagon $ABCDE$, then it must be on one side of the pentagon. Assume that $P$ is on the side $AB$, then it must be the midpoint of $AB$, and $PBCDE$ is a convex pentagon with strictly less area than that of $ABCDE$.
So, the minimum value of $n$ is $11$.