Example 3 Find the largest positive integer $n$, such that in three-dimensional space there exist $n$ points $P_{1}, P_{2}, \cdots, P_{n}$, where no three points are collinear, and for any $1 \leqslant i<j<k \leqslant n, \triangle P_{i} P_{j} P_{k}$ is not an obtuse triangle.
Solution: The conclusion is $n=8$.
Consider the 8 vertices of a cube, it is easy to verify that any three points form a triangle that is not an obtuse triangle.

Next, we prove $n \leqslant 8$. Let $P_{1}, P_{2}, \cdots, P_{n}$ be a set of points satisfying the requirement, and place them in a Cartesian coordinate system, assuming $P_{1}$ is the origin $O$.

If $P_{1}, P_{2}, \cdots, P_{n}$ are not coplanar, let $\Lambda$ be the convex hull of these points. For $2 \leqslant i \leqslant n$, let $\Lambda_{i}=\Lambda+\overrightarrow{O P}_{i}$ (thus, $\Lambda=\Lambda_{1}$). Consider the transformation that enlarges $\Lambda$ by a factor of 2 centered at the origin, and denote the images of $\Lambda$ and $P_{i}$ under this transformation as $\Gamma$ and $Q_{i}$, respectively. It is easy to see that all $\Lambda_{i}, 1 \leqslant i \leqslant n$ are within $\Gamma$ and the volume of $\Gamma$ is 8 times the volume of $\Lambda$. We claim:
(a) Each $P_{i}$ is a vertex of $\Lambda$;
(b) $\Lambda_{i}, \Lambda_{j}(1 \leqslant i < j \leqslant n)$ do not intersect.

Proof of (a): If $P_{2}$ is not a vertex of $\Lambda$, then there exist points $P_{i}, P_{j}$ such that $\angle P_{i} P_{1} P_{j} > 90^{\circ}$, which is a contradiction! If $P_{2}$ is inside the convex hull $\Lambda$, let the line $P_{1} P_{2}$ intersect a face $\pi$ of $\Lambda$ at point $P_{2}^{\prime}$, and let the projection of $P_{1}$ on the plane $\pi$ be $H$. Since $P_{2}^{\prime}$ is inside the plane $\pi$, there exists a vertex $P_{i}$ on the plane $\pi$ such that $\angle P_{i} P_{2}^{\prime} H \geqslant 90^{\circ}$. Therefore,
$$
P_{1} P_{2}^{\prime 2} + P_{i} P_{2}^{\prime 2} = P_{1} H^{2} + H P_{2}^{\prime 2} + P_{i} P_{2}^{\prime 2} \leqslant P_{1} H^{2} + P_{i} H^{2} = P_{1} P_{i}^{2},
$$
so $\angle P_{1} P_{2} P_{i} > \angle P_{1} P_{2}^{\prime} P_{i} \geqslant 90^{\circ}$,
which is a contradiction! Therefore, (a) holds.

For (b), consider the plane $p_{ij}$ passing through $Q_{ij} = \overrightarrow{O P}_{i} + \overrightarrow{O P}_{j}$ and perpendicular to the line segment $P_{i}P_{j}$. All points of $\Lambda_{i}$ are on the same side of $p_{ij}$ as $Q_{i}$ (otherwise, there would be an obtuse triangle), and all points of $\Lambda_{j}$ are also on the same side of $p_{ij}$ as $Q_{i}$. Therefore, (b) holds.

By (a) and (b), the volume of $\bigcup_{i=1}^{n} \Lambda_{i}$ (which is no greater than the volume of $\Gamma$) is the sum of the volumes of all $\Lambda_{i}(1 \leqslant i \leqslant n)$, which is $n$ times the volume of $\Lambda$. Therefore, $n \leqslant 8$ holds.

If $P_{1}, P_{2}, \cdots, P_{n}$ are coplanar, the same argument shows that $n \leqslant 4$ holds.
Therefore, $n \leqslant 8$.
Hence, the maximum value of $n$ is 8.