Problem 1190

First solution. Let the sum $\vec{\sigma}$ of all vectors be different from zero ($|\vec{\sigma}|=s>0$). Introduce a rectangular coordinate system $Oxy$, in which the axis $Ox$ is directed along $\vec{\sigma}$. Let $\vec{a}$ be the longest vector in the set, i.e., it is not shorter than $\vec{b}=\vec{\sigma}-\vec{a}$. Since the $y$-coordinates of the vectors $\vec{a}$ and $\vec{b}$ are equal in magnitude, the $x$-coordinate $a_{x}$ of the vector $\vec{a}$ in magnitude is not less than the $x$-coordinate $b_{x}=s-a_{x}$ of the vector $\vec{b}$. From this, we get that $a_{x} \geq s / 2$. Now, if all vectors in the set are long, then the sum of their $x$-coordinates is not less than $n s / 2 > s$, but this sum is equal to $s$. Contradiction.

Second solution. Denote the given vectors by $a_{k}(k=1, \ldots, n)$, and their sum by $\vec{\sigma}$. By the condition, $\left|\overrightarrow{\boldsymbol{a}_{k}}\right| \geq\left|\vec{\sigma}-\overrightarrow{\boldsymbol{c}_{k}}\right|$. Square this inequality: $\vec{a}_{k}^{k} \geq \vec{\sigma}^{2}-2 \vec{\sigma}_{\boldsymbol{\sigma}} \vec{a}_{\mathbf{k}}+\vec{a}_{\mathbf{k}}^{2}$. Summing such inequalities for all $k$ from 1 to $n$, we get $0 \geq n \vec{\sigma}^{2}-2 \vec{\sigma} \cdot\left(\overrightarrow{a_{1}}+\overrightarrow{a_{2}}+...+\overrightarrow{a_{n}}\right)$, i.e., $0 \geq(n-2) \vec{\sigma}^{2}$. Therefore, $\vec{\sigma}=\vec{\sigma}$.