Four. (50 points) Given a set of 9 points in space
$$
M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\},
$$

where no four points are coplanar. Connect some line segments between these 9 points to form a graph $G$, such that the graph contains no tetrahedron. Question: What is the maximum number of triangles in graph $G$?
Four, first prove a lemma.
Lemma If in a space graph with $n$ points there is no triangle, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$ ( $[x]$ represents the largest integer not exceeding the real number $x$).

Proof Let the $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, among which, the number of edges drawn from point $A_{1}$ is the most (let's say there are $k$ edges) as $A_{1} A_{n}, A_{1} A_{n-1}, \cdots, A_{1} A_{n-k+1}$.

Since there is no triangle, there are no edges between the points $A_{n}, A_{n-1}, \cdots, A_{n-k+1}$.

Thus, each edge in the space graph has at least one endpoint among the points $A_{1}, A_{2}, \cdots, A_{n-k}$, and each $A_{i}(1 \leqslant i \leqslant n-k)$ can draw at most $k$ edges.
Therefore, the total number of edges is less than or equal to $k(n-k) \leqslant\left[\frac{n^{2}}{4}\right]$.
Back to the original problem.
Next, prove: If graph $G$ already has (at least) 28 triangles, then there is at least one tetrahedron.

Otherwise, in the 9-point set $M=\left\{A_{1}, A_{2}, \cdots, A_{9}\right\}$, by the pigeonhole principle, there must be a point that is at least
$$
\left[\frac{28 \times 3}{9}\right]+1=10
$$

vertices of triangles. Thus, at least 5 edges are drawn from this point (let this point be $A_{1}$).
(1) If 5 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{6}$ are drawn from point $A_{1}$, according to the problem, since there is no tetrahedron, the subgraph formed by the 5 points $A_{2}, A_{3}, \cdots, A_{6}$ has no triangle. By the lemma, this subgraph has at most $\left[\frac{5^{2}}{4}\right]=6$ edges. Thus, the number of triangles with $A_{1}$ as a vertex is at most 6, a contradiction.
(2) If 6 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{7}$ are drawn from point $A_{1}$, similarly to (1), the number of triangles with $A_{1}$ as a vertex is at most $\left[\frac{6^{2}}{4}\right]=9$, a contradiction.
(3) If 7 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{8}$ are drawn from point $A_{1}$, since there is no tetrahedron, the subgraph formed by the 7 points $A_{2}, A_{3}, \cdots, A_{8}$ has no triangle. This subgraph has at most $\left[\frac{7^{2}}{4}\right]=12$ edges. Thus, the number of triangles with $A_{1}$ as a vertex is at most 12, and the triangles not having $A_{1}$ as a vertex must have point $A_{9}$ as a vertex. Similarly, there are at most 12 such triangles. Therefore, the total number of triangles is less than or equal to $12 \times 2=24<28$, a contradiction.
(4) If 8 edges $A_{1} A_{2}, A_{1} A_{3}, \cdots, A_{1} A_{9}$ are drawn from point $A_{1}$, the subgraph formed by the 8 points $A_{2}, A_{3}, \cdots, A_{9}$ has no triangle. This subgraph has at most $\left[\frac{8^{2}}{4}\right]=16$ edges. Hence, the graph $G$ has at most 16 triangles, a contradiction.
Thus, the number of triangles satisfying the condition is at most 27.
Divide the 9-point set $M$ into three groups
$$
\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}, A_{6}\right\},\left\{A_{7}, A_{8}, A_{9}\right\},
$$
such that no two points in the same group are connected, while any two points in different groups are connected, thus, there are $C_{3}^{1} C_{3}^{1} C_{3}^{1}=27$ triangles, of course, there is no tetrahedron.