Example 8 In the Cartesian coordinate system, points with integer coordinates for both the x-axis and y-axis are called lattice points, and a convex polygon whose vertices are all lattice points is called a lattice convex polygon. Try to find the minimum number of lattice points inside a lattice convex heptagon.

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First, we prove the following lemma.
Lemma: A lattice convex pentagon $A B C D E$ contains at least one lattice point.
Proof of the Lemma: Without loss of generality, assume that the interiors of the sides of the convex pentagon $A B C D E$ contain no lattice points. In fact, if $D E$ contains a lattice point $F$, and $F A$ contains no lattice points, then we can replace the convex pentagon $A B C D E$ with the lattice convex pentagon $A B C D F$. After a finite number of steps, we can obtain a lattice convex pentagon with no lattice points on the interiors of its sides and within the original lattice convex pentagon $A B C D E$. Thus, we can use this lattice convex pentagon to replace $A B C D E$ for discussion, without affecting the proof of the conclusion.

Since the coordinates of the 5 vertices of the lattice convex pentagon $A B C D E$ have at most four types of parity: (odd, odd), (even, even), (odd, even), (even, odd), there must be two vertices whose coordinates have the same parity. Therefore, the midpoint $M$ of the line segment connecting these two points must be a lattice point. Since the interiors of the sides of the convex pentagon $A B C D E$ contain no lattice points, $M$ must be inside the convex pentagon $A B C D E$. The lemma is proved.
Next, we prove that a lattice convex heptagon $A B C D E F G$ contains at least 4 lattice points.
In fact, by the lemma, the lattice convex pentagon $A B C D E$ contains at least one lattice point $M$. The line $M A$ divides the plane into two sides, and at least 5 of the 7 vertices of the heptagon $A B C D E F G$ must be on one side of the line $M A$. Therefore, at least 3 vertices must be on one side of the line $M A$. Without loss of generality, assume $E, F, G$ are on the same side of $M A$ (as shown in Figure 6-38). Thus, the lattice convex pentagon $A M E F G$ contains at least one lattice point $N$, and the line $M N$ divides the plane into two sides, with at least 3 of the 7 vertices of the heptagon $A B C D E F G$ on one side of the line $M N$. Without loss of generality, assume $A, B, C$ are on the same side of the line $M N$ (as shown in Figure 6-38). By the lemma, the lattice convex pentagon $M N A B C$ contains at least one lattice point $P$. Choose any point $Q$ inside $\triangle P M N$, and draw 3 rays $Q M, Q N, Q P$, dividing the plane into three angular regions: $\angle M Q N, \angle N Q P, \angle P Q M$ (each angular region includes its two sides and the interior of the angle). Thus, the 7 vertices $A, B, C, D, E, F, G$ belong to these 3 angular regions. By the pigeonhole principle, at least $\left[\frac{7-1}{3}\right]+1=3$ of these points belong to the same angular region. Without loss of generality, assume $E, F, G$ belong to $\angle M Q N$ (as shown in Figure 6-38). Thus, the lattice convex pentagon $N M E F G$ contains at least one lattice point $R$. Therefore, the lattice convex heptagon $A B C D E F G$ contains at least 4 lattice points.

On the other hand, as shown in Figure 6-39, the lattice convex heptagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7}$ contains exactly 4 lattice points $A_{8}, A_{9}, A_{10}, A_{11}$.
In summary, a lattice convex heptagon contains at least 4 lattice points.