11. (Mathematics for Middle School, Issue 1, 2005 Olympiad Training Problem) In a plane, there are 12 points, and no three points are collinear. Using any one of these points as the starting point and another as the endpoint, draw all possible vectors. A triangle whose three side vectors sum to the zero vector is called a "zero triangle." Find the maximum number of "zero triangles" that can be formed with these points as vertices.
11. Let these 12 points be $P_{1}, P_{2}, \cdots, P_{12}$. The number of triangles determined by these 12 points is $C_{12}^{3}$. Let the number of vectors starting from $P_{i} (i=1, 2, \cdots, 12)$ be $x_{i}, 0 \leqslant x_{i} \leqslant 11$. If a triangle with 3 points as vertices is a "non-zero triangle", then there is and only one point that is the starting point of the vectors of two sides of this triangle. Therefore, the number of "non-zero triangles" with $P_{i}$ as one of the vertices and the starting point of two sides is $C$ (with the convention $C_{1}^{2}=C_{0}^{2}=0$). Thus, the total number of "non-zero triangles" among the triangles with these points as vertices is $\sum_{i=1}^{12} C_{x_{i}}^{2}$. Therefore, the number of "zero triangles" is 
$C_{12}^{3}-\sum_{i=1}^{12} C_{x_{i}}^{2}$.
First, find the minimum value of $\sum_{i=1}^{12} C_{x_{i}}^{2}$.
Since $\sum_{i=1}^{12} x_{i}=C_{12}^{2}=66$, we have
$\sum_{i=1}^{12} C_{x_{i}}^{2}=\sum_{i=1}^{12} \frac{x_{i}\left(x_{i}-1\right)}{2}=\frac{1}{2} \sum_{i=1}^{12}\left(x_{i}^{2}-x_{i}\right)=\frac{1}{2} \sum_{i=1}^{12} x_{i}^{2}-33$.
Since the non-negative integer $x_{i}$ does not exceed 11, $\sum_{i=1}^{12} x_{i}^{2}$ has a minimum value.
If there exist $1 \leqslant i, j \leqslant 12 (i \neq j)$ such that $x_{i}-x_{j} \geqslant 2$, we can denote
$x^{\prime}{ }_{i}=x_{i}-1, x^{\prime}{ }_{j}=x_{j}{ }^{\prime}+1$.
Clearly, $x_{i}>x_{i}^{\prime} \geqslant x_{j}^{\prime}>x_{j}$, then
$x_{i}^{2}+x_{j}^{2}-\left(x_{i}^{\prime 2}+x_{j}^{\prime 2}\right)=x_{i}^{2}+x_{j}^{2}-\left(x_{i}-1\right)^{2}-\left(x_{j}+1\right)^{2}=2\left(x_{i}-x_{j}-1\right)>0$,
and since $x_{i} \in \mathbf{N}$, for all $1 \leqslant i, j \leqslant 12$, $\sum_{i=1}^{12} x_{i}^{2}$ takes its minimum value only when $\left|x_{i}-x_{j}\right|=0$ or 1, i.e., when
$\left\{x_{1}, x_{2}, \cdots, x_{12}\right\}=\{5,5,5,5,5,5,6,6,6,6,6,6\}$, $\sum_{i=1}^{12} x_{i}^{2}$ takes its minimum value $5^{2} \times 6+6^{2} \times 6=366$.
Therefore, the minimum value of $\sum_{i=1}^{12} C_{x_{i}}^{2}$ is $\frac{1}{2} \times 366-33=150$.
Thus, the maximum number of "zero triangles" is
$C_{12}^{3}-150=70$.
Note In this problem, since $x_{i} \in \mathbf{N}$, the minimum value of $\sum_{i=1}^{12} x_{i}^{2}$ cannot be found using the mean inequality, so this minimum value is not $\frac{6^{2}}{12}=363$.