6. 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?
6. Divide 9 points into 3 groups $\left\{A_{1}, A_{2}, A_{3}\right\},\left\{A_{4}, A_{5}, A_{6}\right\},\left\{A_{7}, A_{8}, A_{9}\right\}$, and connect every two points from different groups. Then these connections do not form a tetrahedron but do form 27 triangles, showing that the maximum number of triangles is at least 27.
The following proof shows: If the graph has 28 triangles, then there must be a tetrahedron.
For 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 the graph contains no triangles, then the maximum number of line segments is $\left[\frac{n^{2}}{4}\right]$.

In fact, suppose that from $A_{1}$, $k$ line segments are drawn, which are $A_{1} B_{1}, A_{1} B_{2}, \cdots, A_{1} B_{k}$. Since there are no triangles, no two points among $B_{1}, B_{2}, \cdots, B_{k}$ are connected. The remaining $n-k-1$ points are $A_{2}, A_{3}, \cdots, A_{n-k}$. Therefore, from $B_{i}$, the number of line segments is at most $n-k$ $(i=1,2, \cdots, k)$, and from $A_{j}$, the number of line segments is at most $k$ $(j=1,2, \cdots, n-k)$. Hence, the number of line segments $t$ in the graph does not exceed $\frac{1}{2}[k(n-k)+(n-k) k]=(n-k) k=\frac{n^{2}}{4}-\left(\frac{n}{2}-k\right)^{2} \leqslant \frac{n^{2}}{4}$.
Since $t$ is an integer, $t \leqslant\left[\frac{n^{2}}{4}\right]$, thus the lemma is proved.
Returning to the proof of the original problem. If there is no tetrahedron, let point $A_{1}$ be the point with the most line segments drawn. Since 28 triangles have 84 vertices, by the pigeonhole principle, there must be a point $A_{2}$, which is a common vertex of at least $\left[\frac{84-1}{9}\right]+1=10$ triangles. If $A_{2}$ draws at most 6 line segments, then the other ends of these lines (at most 6 endpoints) have at least 10 line segments between them, and by the lemma, there must be 3 line segments forming a triangle, plus $A_{2}$ forming a tetrahedron, which is a contradiction. Therefore, point $A_{2}$ must draw at least 7 line segments, and point $A_{1}$ even more so.
(1) If point $A_{1}$ draws 8 line segments, since there is no tetrahedron, the graph formed by the remaining 8 points cannot have a triangle. Thus, by the lemma, the number of line segments among these 8 points does not exceed $\left[\frac{8^{2}}{4}\right]=16$, so the graph can have at most 16 triangles, which is a contradiction.
(2) If point $A_{1}$ draws 7 line segments, then the subgraph formed by the other 7 endpoints of these line segments cannot have a triangle. Thus, by the lemma, there are at most $\left[\frac{7^{2}}{4}\right]=12$ line segments, so the number of triangles containing vertex $A_{1}$ is at most 12. Let the only point among the 9 points that is not connected to $A_{1}$ be $B$. Then, similarly, the number of triangles containing vertex $B$ is at most 12, so the graph can have at most 24 triangles, which is a contradiction. In summary, the graph can have at most 27 triangles.