11. (15 points) Given seven different points on a circle, vectors are drawn from any one point to another (for points $A$ and $B$, if vector $\overrightarrow{A B}$ is drawn, then vector $\overrightarrow{B A}$ is not drawn). If the four sides of a convex quadrilateral determined by any four points are four consecutive vectors, it is called a "zero quadrilateral". Find the maximum number of zero quadrilaterals among the convex quadrilaterals formed by any four of these seven points.
11. Let the seven points on the circumference be $P_{1}, P_{2}, \cdots, P_{7}$. The number of vectors starting from point $P_{i} (i=1,2, \cdots, 7)$ is $x_{i} (i=1,2, \cdots, 7)$, then $0 \leqslant x_{i} \leqslant 6$, and
$$
\sum_{i=1}^{7} x_{i}=\mathrm{C}_{7}^{2}=21 \text {. }
$$

First, find the minimum number of "non-zero quadrilaterals" among the $\mathrm{C}_{7}^{4}=35$ convex quadrilaterals.
It is easy to know that non-zero quadrilaterals only have the following two scenarios:
The first type: A non-zero quadrilateral with exactly one vertex as the starting point of two vectors (as shown in Figure 3).

The second type: A non-zero quadrilateral with exactly two vertices as the starting points of two vectors (as shown in Figure 4).
(1) When the entire figure contains the second type of non-zero quadrilateral, it is easy to know that on the basis of the four vertices of the second type of non-zero quadrilateral, each additional point will increase four convex quadrilaterals compared to the original quadrilateral, and it is easy to know from the drawing that these four convex quadrilaterals are all non-zero quadrilaterals.

Thus, in the 35 convex quadrilaterals in the entire seven-point figure, if there is one second type of non-zero quadrilateral, then on the basis of this non-zero quadrilateral, adding three points will increase at least $(4 \times 3=) 12$ non-zero quadrilaterals. Therefore, at this time, there are at least 13 non-zero quadrilaterals in the figure.
(2) When the seven-point figure does not contain the second type of non-zero quadrilateral, i.e., it only contains the first type of non-zero quadrilateral and zero quadrilateral, the problem can be converted to just finding the minimum number of the first type of non-zero quadrilateral.

And the four vertices of the first type of non-zero quadrilateral have exactly one vertex as the starting point of two vectors of this non-zero quadrilateral, and these two vectors form a diagonal vector. When these three vectors are determined, then a non-zero quadrilateral is determined. Therefore, when the figure does not contain the second type of non-zero quadrilateral, the number of the first type of non-zero quadrilateral in the figure
$$
M \geqslant \sum_{i=1}^{7} \mathrm{C}_{x_{i}}^{3} .
$$

Since $x_{1}+x_{2}+\cdots+x_{7}=21$, $M$ has a minimum value.

If there is some $x_{i}$ in $x_{1}, x_{2}, \cdots, x_{7}$ that is less than 3, then there must be some $x_{j}$ that is greater than or equal to 4.

And $C_{4}^{3}=4, C_{5}^{3}=10, C_{6}^{3}=20$, and $C_{3}^{3}=1$, then $M \geqslant 7$.
If $x_{1}, x_{2}, \cdots, x_{7}$ are all not less than 3, by
$$
x_{1}+x_{2}+\cdots+x_{7}=21
$$

we know, only $x_{1}=x_{2}=\cdots=x_{7}=3, M=7 \mathrm{C}_{3}^{3}=7$.
Below, through drawing, it is shown that the figure with $M=7$ exists (as shown in Figure 5).

Figure 5 contains seven non-zero quadrilaterals.
(1) $P_{1} P_{2} P_{3} P_{4}$, (2) $P_{2} P_{3} P_{4} P_{5}$,
(3) $P_{3} P_{4} P_{5} P_{6}$, (4) $P_{4} P_{5} P_{6} P_{7}$,
(5) $P_{5} P_{6} P_{7} P_{1}$, (6) $P_{6} P_{7} P_{1} P_{2}$,
(7) $P_{7} P_{1} P_{2} P_{3}$.

Therefore, the minimum value of $M$ is 7.
Thus, the maximum number of zero quadrilaterals is $C_{7}^{4}-7=28$.