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I have a counterexample in $\mathbb R^2$. Here's how it goes.

Pick two numbers $n$ and $N$, with $N>>n>>1$.
The collection {$v_i$} consists of:

• $N(n-1)$ times the vector $(\frac{n-2}{n},\frac{1}{N})$(\frac{n-2}{n},\frac{1}{N(2n-3)})$•$N(n-2)$times the vector$(-\frac{n-1}{n},\frac{1}{N})$(-\frac{n-1}{n},\frac{1}{N(2n-3)})$

• The vector $(0,-1)$ once.

The smallest ball into which those vectors can be fit back-to-back has radius diameter $\sqrt{5}-\varepsilon$.

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I have a counterexample in $\mathbb R^2$. Here's how it goes.

Pick two numbers $n$ and $N$, with $N>>n>>1$.
The collection {$v_i$} consists of:

• $N(n-1)$ times the vector $(\frac{n-2}{n},\frac{1}{N})$

• $N(n-2)$ times the vector $(-\frac{n-1}{n},\frac{1}{N})$

• The vector $(0,-1)$ once.

The smallest ball into which those vectors can be fit back-to-back has radius $\sqrt{5}-\varepsilon$.