how prove $\varepsilon_i \in \left \{ -1,1 \right \}$ such $\left | \sum_{i=1}^{n} \varepsilon_i z_i \right | \leqslant 1.$ Let $n\geqslant 3$ be an odd number, and $z_1,z_2,...,z_n$ be complex numbers such that $|z_i|=1$ for all $i$.
Prove that there exist $\varepsilon_i \in \left \{ -1,1 \right \}$ satisfying $$\left | \sum_{i=1}^{n} \varepsilon_i z_i \right | \leqslant 1.$$
for  the right  is $2$ proof can see links,and the right is $\sqrt{3}$ see: links2, I think $1$ also is true
My try: use induction:
since
$$\left|\sum_{i=1}^{n} \varepsilon_i z_i +z_{n+1}\right|^2+\left|\sum_{i=1}^{n} \varepsilon_i z_i -z_{n+1}\right|^2=2(|\left|\sum_{i=1}^{n} \varepsilon_i z_i \right|^2+|z_{n+1}|^2)\le 4$$
so maybe prove this $2$ is very easy,But I can't the right constant is$1$

From the following claim, your bound follows for $n=3$. For larger $n$, I do not know how to prove it.
Claim: Let $z_1,z_2,z_2$ with $|z_i|=1$. Then there exist $\varepsilon _1, \varepsilon _2 \in \{-1,1\}$, such that the distance between $\varepsilon _1 z_1+\varepsilon _2 z_2$ and $z_3$ is less than 1.
Proof: Geometrically, a complex number with $|z|=1$ lies on the unit circle. 
Then the set of corner-points
$$
C=\{z_1+z_2,z_1-z_2,-z_1+ z_2,-z_1 -z_2\}
$$
forms a parallelogram and the midpoints
$$
M=\{z_1,-z_1,z_2,-z_2\}
$$
of the four sides have distance $1$ from some points in $C$. The midpoints $M$ lie on the unit-circle (centered at $0$). All points $z_3$ on a unit-circle-segment (red in picture) between two midpoints have distance $\leq 1$ from one of the corners. Choosing the right corner corresponds to choosing the right $\varepsilon_i$ for $i=1,2$. $\quad \quad \quad \square$ 

