How prove $|\xi_{1}v_{1}+\xi_{2}v_{2}+\cdots+\xi_{n}v_{n}|\le 1$ Question:

Let $n\ge 3\in N^{+}$ be  odd numbers, and $v_{1},v_{2},\cdots,v_{n}$ be vectors in the plane with lengths equal to $1$. Prove that there exist  $\xi_{1},\xi_{2},\cdots,\xi_{n}\in\{-1,1\}$ such that
  $$|\xi_{1}v_{1}+\xi_{2}v_{2}+\cdots+\xi_{n}v_{n}|\le 1$$

I find my problem very similar to this Romanian Mo 2001 problem 4: see:
link
But my problem is hard, and I think is true.  Thank you for you help.
 A: I show below that the claim is true for $n=3$.
We can write $v_k=(\cos(\theta_k),\sin(\theta_k))$ with $\theta_k\in {\mathbb R}$,
for $k=1,2,3$. We may assume without loss that $\theta_1 < \theta_2 < \theta_3$ and
$\theta_3-\theta_1 \leq \pi$ (note that $v_k$ can be replaced with $-v_k$, in other words $\theta_k$ may be replaced with $\pi+\theta_k$). Then, the identity
$$
\big|e^{i\theta_1}-e^{i\theta_2}+e^{i\theta_3}\big|^2=
1-4\cos\bigg(\frac{\theta_3-\theta_1}{2}\bigg)\Bigg(
\cos\bigg(\frac{\theta_3-\theta_1}{2}\bigg)-
\cos\bigg(\theta_2-\frac{\theta_1+\theta_3}{3}\bigg)
\Bigg)
$$
or, if you prefer (thanks to WimC for pointing this out),
$$
\big|e^{i\theta_1}-e^{i\theta_2}+e^{i\theta_3}\big|^2=
1-8\cos\bigg(\frac{\theta_3-\theta_1}{2}\bigg)
\sin\bigg(\frac{\theta_3-\theta_1}{2}\bigg)\sin\bigg(\frac{\theta_2-\theta_1}{2}\bigg)
$$
shows that $\big|\big|v_1-v_2+v_3\big|\big| \leq 1$.
A: For convenience I renumber the vectors $v_0, \ldots, v_{n-1}$. Without loss of generality we can assume that all $\pm v_0, \dots, \pm v_{n-1}$ are different, $v_0 = (1,0)$ and all other $v_k$ have a strictly positive $y$-coordinate.  This situation will be considered below.
Let $n$ be odd and $0=\theta_0< \theta_1<\ldots<\theta_n=\pi$. Extend this sequence by taking $\theta_{-k} = \theta_{n-k}-\pi$ for $k\in\{1,\dots, n\}$. Let $$v_k = (\cos \theta_k, \sin \theta_k) \in \mathbb{R}^2$$ for $k\in\{-n,\ldots,n\}$ and$$v=v_0-v_1+v_2-\ldots+v_{n-1}.$$ I claim that $|v|\leq 1.$ Note that $|v|\leq 1$ if and only if $|\langle w, v\rangle|\leq 1$ for all $w$ on the unit circle. Let $k\in\{0,\dots,n-1\}$, $\alpha \in [\theta_k, \theta_{k+1})$ and $w = (\cos \alpha, \sin \alpha)$. Then $$\langle w, v_k \rangle > \langle w, v_{k-1} \rangle > \ldots > \langle w, v_{k-n+1} \rangle$$ and since $n$ is odd
$$(-1)^k\langle w, v \rangle = \langle w, v_k \rangle - \langle w, v_{k-1} \rangle + \ldots + \langle w, v_{k-n+1} \rangle$$
(alternating signs in the right hand side).  With the strict inequalities above it follows that
$$-1 \leq \langle w, v_{k-n+1} \rangle \leq (-1)^k\langle w, v \rangle \leq \langle w, v_k \rangle \leq 1$$ and so $|\langle \pm w, v \rangle| \leq 1$.  This proves the claim.
