Example 5 Let the set of all integer points (points with integer coordinates) in the plane be denoted as $S$. It is known that for any $n$ points $A_{1}, A_{2}, \cdots, A_{n}$ in $S$, there exists another point $P$ in $S$ such that the segments $A_{i} P(i=1,2, \cdots, n)$ do not contain any points from $S$ internally. Find the maximum possible value of $n$.


The maximum value of the required $n$ is 3.
On one hand, when $n \geqslant 4$, take 4 points $A(0,0), B(1,0), C(0,1)$, $D(1,1)$ in $S$ (the remaining $n-4$ points are chosen arbitrarily), then for any other point $P$ in $S$, analyzing the parity of the coordinates of $P$, there are only 4 cases: (odd, odd), (odd, even), (even, odd), and (even, even). Therefore, among $A, B, C, D$, there is one point such that the midpoint of the line segment connecting it to $P$ is a point in $S$, hence $n \leqslant 3$.

On the other hand, we prove: for any three points $A, B, C$ in $S$, there exists another point $P$ in $S$ such that the interiors of $A P, B P, C P$ do not contain any integer points.

Note that for integer points $X\left(x_{1}, y_{1}\right)$ and $Y\left(x_{2}, y_{2}\right)$ on the plane, the necessary and sufficient condition for the line segment $X Y$ to have no integer points in its interior is $\left(x_{1}-x_{2}, y_{1}-y_{2}\right)=1$. Therefore, we can assume $A\left(a_{1}, a_{2}\right), B\left(b_{1}, b_{2}\right), C(0,0)$.

To find a point $P(x, y)$ such that $A P, B P, C P$ have no integer points in their interiors, $x, y$ need to satisfy the following conditions:
$$\left\{\begin{array}{l}
x \equiv 1(\bmod y) \\
x-a_{1} \equiv 1\left(\bmod y-a_{2}\right) \\
x-b_{1} \equiv 1\left(\bmod y-b_{2}\right)
\end{array}\right.$$

Using the Chinese Remainder Theorem, if there exists $y \in \mathbf{Z}$ such that $y, y-a_{2}, y-b_{2}$ are pairwise coprime, then an $x$ satisfying (2) exists, thus finding a point $P$ that meets the requirements.

Take $y=m\left[a_{2}, b_{2}\right]+1, m \in \mathbf{Z}, m$ to be determined, then $\left(y, y-a_{2}\right)=\left(y, a_{2}\right)=\left(1, a_{2}\right)=1$, similarly $\left(y, y-b_{2}\right)=1$. Therefore, it is only necessary to take $m$ such that
$$\left(y-a_{2}, y-b_{2}\right)=1$$

which is equivalent to
$$\left(y-a_{2}, a_{2}-b_{2}\right)=1$$

To make (3) hold, set $a_{2}=a_{2}^{\prime} d, b_{2}=b_{2}^{\prime} d$, where $a_{2}^{\prime}, b_{2}^{\prime}, d \in \mathbf{Z}$, and $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, then $y=m d a_{2}^{\prime} b_{2}^{\prime}+1$, and thus $y-a_{2}=d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)+1$. Noting that when $\left(a_{2}^{\prime}, b_{2}^{\prime}\right)=1$, $\left(b_{2}^{\prime}, a_{2}^{\prime}-b_{2}^{\prime}\right)=1$, so there exists $m \in \mathbf{Z}$ such that $m b_{2}^{\prime} \equiv 1\left(\bmod a_{2}^{\prime}-b_{2}^{\prime}\right)$, such a determined $m$ satisfies: $d\left(a_{2}^{\prime}-b_{2}^{\prime}\right) \mid d a_{2}^{\prime}\left(m b_{2}^{\prime}-1\right)$, at this time
$$y-a_{2} \equiv 1\left(\bmod a_{2}-b_{2}\right),$$

Therefore, (3) holds.
In summary, the maximum value of $n$ that satisfies the conditions is 3.