Three. (50 points) Given a finite set of planar vectors $M$, for any three elements chosen from $M$, there always exist two elements $\boldsymbol{a}, \boldsymbol{b}$ such that $\boldsymbol{a}+\boldsymbol{b} \in M$. Try to find the maximum number of elements in $M$.


Three, the maximum number of elements in the set $M$ is 7.
Let points $A, B, C$ be any three points on a plane. Consider the 7-element set $M=\{\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C A}, \boldsymbol{B A}, \boldsymbol{C B}, \boldsymbol{A C}, \mathbf{0}\}$, which clearly satisfies the conditions. We will now prove that the number of elements in $M$ cannot exceed 7.

When all elements in $M$ are collinear, move all the starting points to the same point and draw a line $l$ parallel to all the elements. Project all elements of $M$ onto the line $l$. Thus, all vectors in $M$ correspond to coordinates on a number line with the common starting point of the elements in $M$ projected onto $l$ as the origin and an arbitrary direction on $l$ as the positive direction. Therefore, the problem can be transformed into a number set problem (from two-dimensional to one-dimensional).
Next, we prove that there are at most 7 elements in this number set.
First, we prove that there are at most 3 positive numbers in the number set. Assume that there are at least 4 positive elements, and the largest 4 numbers are $a_{1}, a_{2}, a_{3}, a_{4}$, with $a_{1} < a_{2} < a_{3} < a_{4}$. Then, $a_{1} + a_{2} + a_{4} > a_{1} + a_{4} > a_{4}$, so the sum $a_{i} + a_{4} \notin M (i=1,2,3)$. Since there is only one element greater than $a_{3}$, which is $a_{4}$, but $a_{2} + a_{3} > a_{1} + a_{3} > a_{3}$, in the set $\left\{a_{1}, a_{3}, a_{4}\right\}$ or $\left\{a_{2}, a_{3}, a_{4}\right\}$, at least one set has the property that the sum of any two elements is not in $M$. This contradicts the given conditions, so the number set has at most 3 positive numbers. Similarly, the number set has at most 3 negative numbers. Adding a 0, thus, the number set $M$ has at most 7 elements.

When the elements in $M$ are not all collinear, move all the starting points to the same point $O$. By the finiteness of $M$, we can draw a Cartesian coordinate system $x O y$ such that the elements in $M$ are not parallel to the coordinate axes.
Next, we prove that there are at most 3 elements in the upper half-plane.
First, we prove that all elements in the upper half-plane are not collinear. Assume there exist elements $a_{1}$ and $a_{2}$ in the upper half-plane that are collinear. Then, take the element $b$ with the smallest angle with $a_{1}$ and $a_{2}$. Consider the set $\left\{a_{1}, a_{2}, b\right\}$. By the choice of $b$, $a_{1} + b$ and $a_{2} + b$ are not in $M$ (the sum vector lies between the two vectors). Thus, there exists $a_{3}$ in $M$ such that $a_{3} = a_{1} + a_{2}$, so $a_{3}$ is collinear with $a_{1}$ and $a_{2}$. Consider the set $\left\{a_{1}, a_{3}, b\right\}$, similarly, there exists $a_{4}$ collinear with $a_{1}$ and $a_{2}$. Continuing this discussion, there are infinitely many elements in $M$ collinear with $a_{1}$ and $a_{2}$, which is a contradiction. Therefore, all elements in the upper half-plane are not collinear.

Next, we prove that there are at most 3 elements in the upper half-plane. Assume there are at least 4 elements in the upper half-plane. Take 4 adjacent elements $a, b, c, d$ in the upper half-plane in a counterclockwise direction. Consider the set $\{a, b, c\}$, then $b = a + c$; consider the set $\{b, c, d\}$, then $c = b + d$. Thus, $b = a + c = a + b + d$, which implies $a + d = 0$. This contradicts the fact that $a$ and $d$ are in the upper half-plane, so there are at most 3 elements in the upper half-plane. Similarly, there are at most 3 elements in the lower half-plane. Adding the zero vector, thus, the set $M$ has at most 7 elements.
In conclusion, the maximum number of elements in the set $M$ is 7.