Four. (50 points) There are 12 points on a plane, and no three points are collinear. Using any one of these points as the starting point and another point 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 12 points as vertices.
Let the 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 $x_{i}$ $\left(0 \leqslant x_{i} \leqslant 11\right)$.

If a triangle with three points as vertices is a non-zero triangle, then there is and only one point that is the starting point of the two sides vectors of this triangle. Therefore, the number of non-zero triangles with $P_{i}$ as one of the vertices and as the starting point of two sides is $\mathrm{C}_{x_{i}}^{2}$. Thus, the total number of non-zero triangles with these points as vertices is $\sum_{i=1}^{12} c_{x_{i}}^{2}$.
Therefore, the number of zero triangles is $\mathrm{C}_{12}^{3}-\sum_{i=1}^{12} \mathrm{C}_{\mathrm{x}_{i}}^{2}$.
First, find the minimum value of $\sum_{i=1}^{12} C_{x_{i}}^{2}$.
Since $\sum_{i=1}^{12} x_{i}=\mathrm{C}_{12}^{2}=66$, we have
$$
\begin{array}{l}
\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}=\sum_{i=1}^{12} \frac{x_{i}\left(x_{i}-1\right)}{2} \\
=\frac{1}{2} \sum_{i=1}^{12}\left(x_{i}^{2}-x_{i}\right)=\frac{1}{2} \sum_{i=1}^{12} x_{i}^{2}-33 .
\end{array}
$$

And the non-negative integer $x_{i}$ does not exceed 11, so $\sum_{i=1}^{12} x_{i}^{2}$ has a minimum value.
By the adjustment method, when
$$
\left|x_{i}-x_{j}\right|=0,1(1 \leqslant i<j \leqslant 12),
$$

that is, when
$$
\begin{array}{l}
\left\{x_{1}, x_{2}, \cdots, x_{12}\right\} \\
=\{5,5,5,5,5,5,6,6,6,6,6,6\}
\end{array}
$$

$\sum_{i=1}^{12} x_{i}^{2}$ takes the minimum value 366.
Thus, the minimum value of $\sum_{i=1}^{12} \mathrm{C}_{x_{i}}^{2}$ is
$$
\frac{1}{2} \times 366-33=150 \text {. }
$$

Therefore, the maximum number of zero triangles with these 12 points as vertices is 70.