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

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 tetrahedra. How many triangles can graph $G$ have at most?
Three, first prove the following conclusion:
In a space graph with $n$ points, if there are no triangles, then the number of edges does not exceed $\left[\frac{n^{2}}{4}\right]$.

Proof: Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the most, say there are $k$ edges: $A_{1} A_{n}, A_{1} A_{n-1}$, $\cdots, A_{1} A_{n-k+1}$. According to the condition, there are no triangles, so there are no edges between the points $A_{n}$, $A_{n-1}, \cdots, A_{n-k+1}$. Therefore, 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 have at most $k$ edges. Thus, the total number of edges is less than or equal to $k(n-k) \leqslant\left[\frac{n^{2}}{4}\right]$.

Next, prove that in a 9-point set $M$, if no 4 points are coplanar, and several line segments are connected between these 9 points, if the graph $G$ already has (at least) 28 triangles, then there is at least one tetrahedron.
Use proof by contradiction.
Assume there is no tetrahedron in the 9-point set $M=\left\{A_{1}\right.$, $\left.A_{2}, \cdots, A_{9}\right\}$. By the pigeonhole principle, there must be a point that is a vertex of at least $\left[\frac{28 \times 3}{9}\right]+1=10$ 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 condition, since there is no tetrahedron, the subgraph formed by the points $A_{2}, A_{3}, \cdots$, $A_{6}$ has no triangles. By the previous conclusion, this subgraph has at most $\left[\frac{5^{2}}{4}\right]=6$ edges. Therefore, the number of triangles with $A_{1}$ as a vertex is at most 6, which is 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 $\left[\frac{6^{2}}{4}\right]=9$, which is 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 points $A_{2}, A_{3}, \cdots, A_{8}$ has no triangles. This subgraph has at most $\left[\frac{7^{2}}{4}\right]=12$ edges. Therefore, 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, so the total number of triangles is less than or equal to $12 \times 2=24<28$, which is 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 points $A_{2}, A_{3}, \cdots, A_{9}$ has no triangles. By the previous conclusion, this subgraph has at most $\left[\frac{8^{2}}{4}\right]=16$ edges. Therefore, the original graph $G$ has at most 16 triangles, which is a contradiction.
Thus, the number of triangles that meet the requirements is at most 27.

Divide the 9-point set $M$ into three groups $\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}\right.$, $\left.A_{6}\right\},\left\{A_{7}, A_{8}, A_{9}\right\}$, such that no two points in the same group are connected, while any two points in different groups are connected. This way, there are $\mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1} \mathrm{C}_{3}^{1}=27$ triangles, of course, with no tetrahedron.