Problem 2. Determine all natural numbers $n, n \geq 3$, for which there exists a set $M$, consisting of $n$ different, non-zero vectors with equal magnitudes such that $\sum_{\vec{u} \in M} \vec{u}=\overrightarrow{0}$, knowing that for any $\vec{v}, \vec{w} \in M$ we have $\vec{v}+\vec{w} \neq \overrightarrow{0}$.
Solution and grading: The case where $n$ is odd is suitable. For example, consider a regular polygon $A_{1} A_{2} \ldots A_{n}$. Then the set $M_{n}=\left\{\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{2} A_{3}}, \ldots, \overrightarrow{A_{n-1} A_{n}}, \overrightarrow{A_{n} A_{1}}\right\}$ satisfies the conditions. .... $\mathbf{p p}$

We prove that $n$ cannot be equal to 4. Suppose the contrary. Then $M=\{\vec{u}, \vec{v}, \vec{w}, \vec{z}\}$. Let $O$ be a point in the plane. Then there exist $U, V, W, Z$ such that $\vec{u}=\overrightarrow{O U}, \vec{v}=\overrightarrow{O V}, \vec{w}=\overrightarrow{O W}$, and $\vec{z}=\overrightarrow{O Z}$. Since the vectors have the same magnitude, we deduce that these four points are the vertices of an inscribed quadrilateral. Without loss of generality, let $U V Z W$ be this quadrilateral. Let $P, Q$ be the reflections of $V$ and $Z$ across $O$. From $\overrightarrow{O U}+\overrightarrow{O V}+\overrightarrow{O W}+\overrightarrow{O Z}=\overrightarrow{0}$ we get $\overrightarrow{O U}+\overrightarrow{O W}=\overrightarrow{O P}+\overrightarrow{O Q}$, so $U P W Q$ is a parallelogram, possibly degenerate, hence a rectangle. Then $\overrightarrow{O U}+\overrightarrow{O W}=\overrightarrow{0}$, which is not acceptable.

Next, we prove that any even natural number greater than or equal to 6 is suitable. Consider a set of points $\left\{A_{1}, A_{2}, \ldots, A_{n}\right\}$ such that $A_{1} A_{2} A_{3}$ is an equilateral triangle, $A_{3} A_{4} \ldots A_{n}$ is a regular polygon with $n-3$ sides of the same length as the previous triangle, and the measure in radians of the angle $A_{2} A_{3} A_{n}$ is of the form $\alpha \cdot \pi$, where $\alpha$ is an irrational number (for example, the measure of the angle can be $\left.\frac{\pi}{\sqrt{2}}\right)$. Then the set $M=\left\{\overrightarrow{A_{1} A_{2}}, \overrightarrow{A_{2} A_{3}}, \overrightarrow{A_{3} A_{1}}, \overrightarrow{A_{3} A_{4}}, \ldots, \overrightarrow{A_{n-1} A_{n}}, \overrightarrow{A_{n} A_{3}}\right\}$ satisfies the condition of the problem, the irrationality of $\alpha$ ensuring that there are no vectors in the same direction. In conclusion, $n \in \mathbb{N} \backslash\{0,1,2,4\}$.

Note: Alternatively, it can be shown for $n$ odd and $n=4$ as above. Then it is shown that if $n$ is suitable, then $n+3$ is suitable. For this, let $n$ vectors that satisfy the conditions $\overrightarrow{O A_{1}}, \overrightarrow{O A_{2}}, \ldots, \overrightarrow{O A_{n}}$. The lines $O A_{1}, O A_{2}, \ldots, O A_{n}$ together with the $2 n$ lines that form angles of $\frac{\pi}{3}$, respectively $\frac{2 \pi}{3}$ with the first $n$ lines, intersect the circle on which the points $A_{1}, A_{2}, \ldots, A_{n}$ lie in at most $6 n$ points. Choose a point $A_{n+1}$ different from these, situated on the circle, and consider the equilateral triangle $A_{n+1} A_{n+2} A_{n+3}$ inscribed in this circle. The vectors $\overrightarrow{O A_{1}}, \overrightarrow{O A_{2}}, \ldots, \overrightarrow{O A_{n+3}}$ are $n+3$ vectors that satisfy the conditions. For $n \geq 6$ even, we have that $n-3 \geq 3$ is odd and suitable, so $n$ is suitable.