2. Find all positive integers $n$ $(n \geqslant 3)$ such that there exists a set $M$ with $n$ elements, where the elements are distinct non-zero vectors of equal length, and the following conditions are satisfied: $\sum_{u \in M} u=0$, and for any $v, w \in M$, $\boldsymbol{v}+\boldsymbol{w} \neq \mathbf{0}$.
2. First, for any odd number $n$ not less than 3, take $M$ as the $n$ distinct complex roots of the equation $z^{n}-1=0$. Clearly, the set $M$ satisfies the requirements.
Next, consider even numbers $n$ not less than 6.
Let $\frac{n}{2}=k$, first consider the decomposition of $\frac{1}{2}$.
From $\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}$, we know that $\frac{1}{2}$ can be decomposed into the sum of $k-1$ distinct unit fractions.
Assume $\frac{1}{2}=a_{1}+a_{2}+\cdots+a_{k-1}$, first take $k$ unit vectors above the $x$-axis, whose x-coordinates are sequentially $\frac{1}{2},-a_{1}, \cdots,-a_{n}$, then reflect these vectors about the $x$-axis to get $n$ vectors. Let this set of $n$ vectors be $M$, clearly the set $M$ satisfies the conditions.
Finally, prove that $n=4$ does not satisfy the conditions.
Assume the four vectors are $(1,0),\left(x_{1}, y_{1}\right)$, $\left(x_{2}, y_{2}\right),\left(x_{3}, y_{3}\right)$. To satisfy the conditions, we need $y_{i} \neq 0$ $(i=1,2,3)$.
Assume $y_{1}>0, y_{2}>0, y_{3}<0$. At this point,
$$
\begin{array}{l}
x_{1}+x_{2}+x_{3}=-1, \\
y_{1}+y_{2}+y_{3}=0, \\
x_{i}^{2}+y_{i}^{2}=1(i=1,2,3) . \\
\text { Then }\left\{\begin{array}{l}
\sqrt{1-x_{1}^{2}}+\sqrt{1-x_{2}^{2}}=-y_{3}, \\
x_{1}+x_{2}=-1-x_{3} .
\end{array}\right. \\
(1)^{2}+(2)^{2} \text { gives } \\
x_{1} x_{2}+\sqrt{1-x_{1}^{2}-x_{2}^{2}+x_{1}^{2} x_{2}^{2}}=x_{3} . \\
\text { Let }\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{1} x_{2}=b .
\end{array}\right.
\end{array}
$$

Substituting into equation (3) gives
$$
\begin{array}{l}
b+\sqrt{1-a^{2}+2 b+b^{2}}=-a-1 \\
\Rightarrow a(a+b+1)=0 .
\end{array}
$$

If $a=0$, then $x_{3}=-1$, at this time, $y_{3}=0$, which is a contradiction.
If $a+b+1=0$, it can be deduced that
$$
\left(x_{1}+1\right)\left(x_{2}+1\right)=0 \text {, }
$$

At this time, $y_{1}$ or $y_{2}$ must be 0, which is a contradiction.
In summary, the solution is $\{n \in \mathbf{N} \mid n \geqslant 3, n \neq 4\}$.