How prove this the number of ordered $n$-tuples $(\varepsilon_{1},\cdots,\varepsilon_{n})$such this following inequality is $2^{n-100}$ Interesting Question:

for any complex numbers $z_{1},z_{2},\cdots,z_{n}$ such
  $$\begin{cases}
|z_{1}|^2+|z_{2}|^2+\cdots+|z_{n}|^2=1\\
|z_{i}|\le\dfrac{1}{10},i=1,2,\cdots,n
\end{cases}$$
   show that the number of ordered $n$-tuples $(\varepsilon_{1},\varepsilon_{2},\cdots,\varepsilon_{n})$such this following inequality
  $$|z_{1}\varepsilon_{1}+z_{2}\varepsilon_{1}+\cdots+z_{n}\varepsilon_{n}|\le\dfrac{1}{3}$$
  at least $2^{n-100}$

where $\varepsilon_{1},\varepsilon_{2},\cdots,\varepsilon_{n}\in \{-1,1\}$
This problem is from china 2014 omlypiad problem exisice it,and I find sometimes with this background:
(2004 Romania )Prove that for any complex numbers $z_{1},z_{2},\cdots,z_{n}$ satisfying
$|z_{1}|^2+|z_{2}|^2+\cdots+|z_{n}|^2=1$, one can selcet
 $\varepsilon_{1},\varepsilon_{2},\cdots,\varepsilon_{n}\in \{-1,1\}$ such that
$$\left|\sum_{k=1}^{n}\varepsilon_{k}z_{k}\right|\le 1$$
then I found Nearest IMRE B´ AR´ ANY, BORIS reseacher these some problem:
see:http://arxiv.org/abs/1303.2877
and
http://arxiv.org/abs/1310.0910  these two paper have some new reslut,but I read find see can't solve my problem,can someone help.Thank you
 A: Partition $\{1,\dots,n\}$ into thirty sets $I_1,\dots,I_{30}$ such that $\sum_{i\in I_j}|z_i|^2<\tfrac1{30}+\tfrac1{100}<\tfrac4{90}$ for each $1\leq j\leq 30.$ This can be achieved by starting with empty sets and adding greedily: the only impediment to adding a value to some $I_j$ is that $\sum_{i\in I_j}|z_i|^2\geq \tfrac1{30},$ but if that holds for all the $j$ then all the values must already be assigned.
Assume these lemmas for now:
Lemma 1. For any positive integer $N$ and any values $z_1,\dots,z_N$ with $\sum_{i=1}^N |z_i|^2\leq 4/90,$ there exist at least $2^N/10$ tuples $(\epsilon_1,\dots,\epsilon_N)\in\{-1,1\}^{N}$ such that $|\sum_{i=1}^{N}\epsilon_iz_i|^2\leq 5/90$ and $\epsilon_1=1.$
Lemma 2. Given values $z_1,\dots,z_N$ such that $|z_i|^2\leq 5/90$ for $1\leq i\leq N,$ there exist $\epsilon_1,\dots,\epsilon_N\in\{-1,1\}$ such that $|\sum_{i=1}^N\epsilon_iz_i|\leq 1/3.$ (Actually we only need the case $N=30.$)
Combining the tuples given by applying Lemma 1 to each $I_j,$ there are at least $2^n10^{-30}=2^n1000^{-10}>2^n1024^{-10}=2^{n-100}$ tuples $(\epsilon_1,\dots,\epsilon_n)\in\{-1,1\}^n$ such that $|\sum_{i\in I_j} \epsilon_iz_i|^2\leq 5/90$ and $\epsilon_{\min(I_j)}=1$ for each $j.$
For each of these tuples, by Lemma 2, we can find $(\theta_1,\dots,\theta_{30})\in\{-1,1\}^{30}$ such that $|\sum_{j=1}^{30}\theta_j(\sum_{i\in I_j} \epsilon_iz_i)|\leq 1/3.$ This gives a new tuple defined by $\epsilon'_i=\epsilon_i\theta_j$ for each $i\in I_j$ and $1\leq j\leq 30.$ The resulting $2^{n-100}$ tuples $\epsilon'$ are distinct and satisfy $|\sum_{i=1}^{n}\epsilon'_iz_i|\leq 1/3$ as required. 
Proof of Lemma 1:
$$\sum_{\epsilon\in\{-1,1\}^N}\sum_{i=1}^{N}|\epsilon_iz_i|^2
=\sum_{\epsilon\in\{-1,1\}^N}\sum_{i=1}^{N}\epsilon_iz_i\overline{\sum_{j=1}^{N}\epsilon_jz_j}
=2^N\sum_{i=1}^{N}|z_i|^2\leq\tfrac{4}{90}2^N$$ because the $z_iz_j$ terms cancel. So it is not possible for more than $\tfrac45 2^{N}$ tuples $\epsilon$ to satisfy $|\sum_{i=1}^{N}\epsilon_iz_i|^2>5/90.$ (This is a form of Markov's inequality.) So at least $\tfrac15 2^{N}$ must have $|\sum_{i=1}^{N}\epsilon_iz_i|^2\leq 5/90.$ Requiring $\epsilon_1=1$ halves this to $\tfrac1{10} 2^{N}.$
Proof of Lemma 2: Given any three complex values I claim there are always two $z,w$ such that $|z+w|$ or $|z-w|$ is at most $\max(|z|,|w|).$ This follows from the fact that some two make an angle of at most $\pi/3$ after possibly negating some values: assume one value $z$ lies on the real axis, then $z$ or $-z$ makes an angle of less than $\pi/3$ which anything not in the region $\arg w\in(\pi/3,2\pi/3)\cup (4\pi/3,5\pi/3)$; but any two points in this region make an angle of at most $\pi/6$ with each other after possibly negating one. In algebraic terms this means $2\mathrm{Re}(z\overline w)\geq \tfrac12 |z||w|,$ giving $|z-w|^2=|z|^2+|w|^2-2\mathrm{Re}(z\overline w)\leq\max(|z|,|w|).$
This lets us reduce to the case $N=2.$ Without loss of generality $z_1$ is a positive real, and negating $z_2$ if necessary we can assume $\mathrm{Re}(z_2)\leq 0.$ This gives $|z_1+z_2|^2\leq|z_1|^2+|z_2|^2\leq 10/90$ as required.
