Lemma 1:

If the angle, $\theta$, between two 'short' vectors, $v_1$ and $v_2$ (placed head to tail) satisfies $\theta \leq \pi/3$, then their sum is a short vector.

Proof:

Place $v_1$ at the origin, then the terminal point of $v_1$ lies within the unit ball. Placing the $v_2$ at the tail of $v_1$ creates an angle $\theta \leq \pi/3$. Any arc of radius $r < 1$ traced between $\pi/3$ and $-\pi/3$ from the terminal point of $v_1$ lies within the unit ball as well. Thus $v_1+v_2$ is a 'short' vector. $\blacksquare$

Since $\Sigma v_i = 0$ we can order the vectors in such a way as to create a convex polygon. If any of the interior angles satisfy the conditions for lemma 1, then we can reduce the problem to a polygon with fewer sides.

Lemma 2:

In a convex n-gon (n=4,5 or 6), with interior angles $\theta_i$, either $\theta_i \leq \pi/3$ for some $i$ or at least one pair of adjacent sides (as vectors) may be interchanged so that an angle less than $\pi/3$ is created.

Proof:

If $v_1$ and $v_2$ create interior angle $\alpha$ and $v_2$ and $v_3$ create angle $\beta$, then interchanging $v_2$ and $v_3$ creates an angle $\alpha+\beta-\pi$ between $v_1$ and $v_3$ (which are now adjacent by exchanging $v_2$ and $v_3$). Aiming for a contradiction, assume that this new angle, $\alpha+\beta-\pi > \pi/3$. Thus, $\alpha+\beta > 4\pi/3$. If this were true for every pair of adjacent angles in a quadrilateral, then $16\pi/3 < 2(\Sigma^4_{i=1} \theta_i)$. $\sharp$ For a 5-gon, $20\pi/3 < 2(\Sigma^5_{i=1} \theta_i)$. $\sharp$ And for a 6-gon $8\pi < 2(\Sigma^6_{i=1} \theta_i)$. $\sharp$ Thus, there exists at least one pair of adjacent vectors which may be exchanged to get an interior angle less than or equal to $\pi/3$. $\blacksquare$

Thus, in a 6-gon, we can always exchange two vectors to get the sum of two adjacent vectors to be 'short'. Replacing $v_1, ... ,v_5, v_6$ with $v_1, ... , v_4, v_5+v_6$ to get 5 'short' vectors whose sum is zero. If $v_5+v_6$ is pair-able with another vector in such a way that their sum is 'short', then we are done. Otherwise, we get two other vectors whose sum is a 'short' vector, and we reduce to the four vector case. Because we can again reduce the quadrilateral case, we are done unless we must pair the two non-"sum" vectors to get a 'short' vector. If this is the case, I claim we still have the sum of three of the original six vectors is 'short'.

Proof of claim:

Looks like fedja just beat me to it, but this was just too much to write to delete it. :P