$4 \cdot 20$ Given 9 points in space, where no 4 points are coplanar. Connect several line segments between the 9 points so that there is no tetrahedron in the graph. How many triangles can there be at most in the graph?
[Solution] Divide 9 points into 3 groups: $\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}, A_{6}\right\},\left\{A_{7}, A_{8}\right.$, $\left.A_{9}\right\}$, and connect a line between every two points that belong to different groups. Then these lines do not form a tetrahedron but do form 27 triangles. It is evident that the maximum number of triangles is no less than 27.

The following proof shows that if the graph has 28 triangles, then there must be a tetrahedron. To prove this, we first prove the following lemma.

Lemma: Given $n$ points in space, where no 4 points are coplanar. If several line segments are connected and there are no triangles in the graph, then the number of line segments is at most $\left[\frac{n^{2}}{4}\right]$.

Proof of the Lemma: We use mathematical induction to prove that when $n$ points are connected by $\left[\frac{n^{2}}{4}\right]+1$ line segments, these line segments must form at least 1 triangle. When $n=3,4$, the proposition is obviously true. Assume the proposition holds for $n=2 k-1$. When $n=2 k$, $\left[\frac{n^{2}}{4}\right]+1=k^{2}+1$, i.e., there are $k^{2}+1$ line segments between $2 k$ points, which have $2 k^{2}+2$ endpoints. Since
$$
\frac{2 k^{2}+2}{2 k}=k+\frac{1}{k},
$$
it follows that there must be a point $A$ that connects to at most $k$ line segments. Removing point $A$, the remaining $2 k-1$ points have at least
$$
k(k-1)+1=\left[\frac{(2 k-1)^{2}}{4}\right]+1
$$
line segments. By the induction hypothesis, they must form at least 1 triangle, i.e., the proposition holds for $n=2 k$. When $n=2 k+1$, $\left[\frac{(2 k+1)^{2}}{4}\right]+1=k(k+1)+1$, i.e., there are $k(k+1)+1$ line segments between $2 k+1$ points, which have $2 k(k+1)+2$ endpoints. Since
$$
\frac{2 k(k+1)+2}{2 k+1}=k+\frac{k+2}{2 k+1},
$$
it follows that there must be a point $A$ that connects to at most $k$ line segments. Removing point $A$, the remaining $2 k$ points have at least $k^{2}+1$ line segments. Thus, they must form at least 1 triangle, i.e., the proposition holds for $n=2 k+1$. Therefore, by induction, the proposition holds for all $n \geqslant 3$.

Returning to the proof of the original problem. If there is no tetrahedron, let point $A_{1}$ be the point that connects to the most line segments. Since 28 triangles have 84 vertices, by the pigeonhole principle, there must be a point $A_{2}$ that is a common vertex of at least 10 triangles. If point $A_{2}$ connects to at most 6 line segments, then these lines have at most 6 other endpoints, and there are at least 10 line segments among these endpoints. By the lemma, there must be a triangle, and adding point $A_{2}$ forms a tetrahedron, which is a contradiction. Therefore, point $A_{2}$ must connect to at least 7 line segments, and point $A_{1}$ must connect to even more.
(1) If point $A_{1}$ connects to 8 line segments, then since there is no tetrahedron, the subgraph formed by the remaining 8 points cannot have a triangle. By the lemma, there can be at most 16 line segments among these 8 points, so the graph can have at most 16 triangles, which is a contradiction.
(2) If point $A_{1}$ connects to 7 line segments, then the subgraph formed by the other 7 endpoints of these 7 line segments cannot have a triangle. By the lemma, there can be at most 12 line segments among these 7 points. Thus, the graph can have at most 24 triangles, which is a contradiction.
In conclusion, the graph can have at most 27 triangles.