Given 9 points in space, where any 4 points are not coplanar. Connect several line segments among these 9 points so that there is no tetrahedron in the graph. How many triangles can there be at most in the graph?

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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 $\left[\frac{n^{2}}{4}\right]$.

Let these $n$ points be $A_{1}, A_{2}, \cdots, A_{n}$, where the number of edges from $A_{1}$ is the maximum, say $k$ 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 $A_{n-k+1}, \cdots, A_{n}$. 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 have at most $k$ edges, so the total number of edges $\leqslant k(n-k) \leqslant \left(\frac{n^{2}}{4}\right)$.

Next, prove that in a space with 9 points, where no 4 points are coplanar, and several line segments are connected between these 9 points, if the graph contains at least 28 triangles, then there is at least one tetrahedron.

Use proof by contradiction. Assume there is no tetrahedron, and let these 9 points be $A_{1}, A_{2}, \cdots, A_{9}$. By the pigeonhole principle, there must be a point that is a vertex of at least $\left\{\frac{28 \times 3}{9}\right\} = 10$ triangles, where $\left\{\frac{28 \times 3}{9}\right\}$ represents the smallest integer not less than $\frac{28 \times 3}{9}$. Thus, at least 5 edges are drawn from this point, say $A_{1}$.
(1) If 5 edges $A_{1} A_{2}, \cdots, A_{1} A_{6}$ are drawn from $A_{1}$, then by the condition, since there is no tetrahedron, the subgraph formed by the points $A_{2}, A_{3}, \cdots, A_{6}$ has no triangles. By conclusion (*), this subgraph has at most $\left[\frac{5^{2}}{4}\right] = 6$ edges, so 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 $A_{1}$, then similarly, 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 $A_{1} A_{2}, \cdots, A_{1} A_{8}$ are drawn from $A_{1}$, since there is no tetrahedron, the subgraph formed by the points $A_{2}, \cdots, A_{8}$ has no triangles. Using conclusion (*), 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 number of triangles not having $A_{1}$ as a vertex must have $A_{9}$ as a vertex, and similarly, there are at most 12 such triangles. Therefore, the total number of triangles $\leqslant 12 \times 2 = 24 < 28$, which is a contradiction.
(4) If 8 edges $A_{1} A_{2}, \cdots, A_{1} A_{9}$ are drawn from $A_{1}$, then the subgraph formed by the points $A_{2}, \cdots, A_{9}$ has no triangles. By conclusion (*), this subgraph has at most $\left[\frac{8^{2}}{4}\right] = 16$ edges, so the original graph has at most 16 triangles, which is a contradiction.
Thus, the number of triangles satisfying the conditions is at most 27.

Divide the 9 points $A_{1}, A_{2}, \cdots, A_{9}$ into 3 groups $\left\{A_{1}, A_{2}, A_{3}\right\}$, $\left\{A_{4}, A_{5}, A_{6}\right\}$, $\left\{A_{7}, A_{8}, A_{9}\right\}$, with no edges between points in the same group, and edges between points in different groups. This way, there are $C_{3}^{1} C_{3}^{1} C_{3}^{1} = 27$ triangles. Of course, there is no tetrahedron.