i will write the proof to this later...

Lemma: For any $n$ complex numbers on the unit circle we have: $\sum_{1\leqslant j<k\leqslant n}{\lvert \omega_j-\omega_k\rvert}^2=n^2-{\left\lvert\sum_{j=1}^{n}\omega_j\right\rvert}^2$ Proof: $$ \sum_{1\leqslant j<k\leqslant n}{\lvert \omega_j-\omega_k\rvert}^2=\sum_{1\leqslant j<k\leqslant n}(\omega_j-\omega_k)\overline{(\omega_j-\omega_k)}=(n-1)\sum_{j=1}^{n}\omega_j\overline{\omega_j}+\sum_{1\leqslant j<k\leqslant n}(\omega_j\overline{\omega_k}+\overline{\omega_j}\omega_k)$$$${\left\lvert\sum_{j=1}^{n}\omega_j\right\rvert}^2=(\sum_{j=1}^{n}\omega_j)(\sum_{j=1}^{n}\overline{\omega_j})=n+\sum_{1\leqslant j<k\leqslant n}(\omega_j\overline{\omega_k}+\overline{\omega_j}\omega_k)$$And subtracting the two preceding equations proves the lemma. Assume that $\omega_1,\omega_2,\cdots,\omega_n$ exist with the problem's condition. The lemma implies that: $$\sum_{1\leqslant i<j\leqslant k}{\lvert \omega_j-\omega_k\rvert}^2=\sum_{1\leqslant i<j\leqslant k}{\lvert \omega_j^2-\omega_k^2\rvert}^2=2n-1$$$$\implies\sum_{1\leqslant i<j\leqslant k}{\lvert \omega_j-\omega_k\rvert}^2{\lvert \omega_j+\omega_k\rvert}^2=\sum_{1\leqslant i<j\leqslant k}{\lvert \omega_j-\omega_k\rvert}^2$$Assume $a$, $b$ such that $\lvert\omega_a+\omega_b\rvert$ be minimum. Because all of the $w_j$'s are not equal, we have $\lvert \omega_a+\omega_b\rvert\leqslant1$. And we have: $$n-1=\left\lvert\sum_{j=1}^{n}\omega_j\right\rvert\leqslant\lvert\omega_a+\omega_b\rvert+\sum_{j\not\in\{a,b\}}\lvert\omega_j\rvert=n-2+\lvert\omega_a+\omega_b\rvert\implies\lvert\omega_a+\omega_b\rvert\geqslant1$$So the equality case occurs and $\{\omega_j\}_{j\not\in\{a,b\}}$s are on the internal angle bisector of $\omega_a$ , $\omega_b$. [asy][asy] pair a=dir(17); pair b=dir(137); pair c=dir(77); draw(unitcircle); dot("$\omega_a$",a,dir(a)); dot("$\omega_b$",b,dir(b)); dot("All $\{\omega_j\}_{j\not\in\{a,b\}}$s are here",c,dir(c)); [/asy][/asy] After observation $\omega_j^2$s we see that all $\{\omega_j^2\}_{j\not\in\{a,b\}}$s are on the external angle bisector of $\omega_a^2$ , $\omega_b^2$. [asy][asy] size(170); pair a=dir(34); pair b=dir(274); pair c=dir(134); draw(unitcircle); dot("$\omega_a^2$",a,dir(a)); dot("$\omega_b^2$",b,dir(b)); dot("All $\{\omega_j^2\}_{j\not\in\{a,b\}}$s are here",c,N); [/asy][/asy] But we also have: $$n-1=\left\lvert\sum_{j=1}^{n}\omega_j^2\right\rvert\leqslant\lvert\omega_a^2+\omega_b^2\rvert+\sum_{j\not\in\{a,b\}}\lvert\omega_j^2\rvert=n-1$$So again the equality holds and $\{\omega_j^2\}_{j\not\in\{a,b\}}$s should be on the internal angle bisector of $\omega_a^2$ , $\omega_b^2$. So, $\{1,2,\ldots,n\}\setminus\{a,b\}=\emptyset\implies n=2$ And the equality case is when $n=2\quad\wedge\quad\frac{\omega_a}{\omega_b}\in\{cis(\frac{2\pi}{3}),cis(\frac{-2\pi}{3})\}$.

Lemma 3.0 It's possible to assume that $\omega_i$s are all in the semi plane $\textbf{Im}(z)\ge0$ and also one of the $\omega_i$'s like $\omega_1$ is equal to 1. It's possible to multiplicate all $\omega_i$s to a number like $\textbf{cis}(\theta)$ that the real part of $\sum\omega_i$s becomes zero. So we have $$\sum\omega_k=(n-1)i$$therefore $$\sum\textbf{Im}(\omega_k)=n-1$$$\textbf{Im}(\omega_k)$s are real numbers in the interval $[-1,1]$. So if one of the numbers is negative the summation of other $n-1$ numbers is less than or equal to $n-1$ and therefore the summation of all of them becomes less than $n-1$ that's a contradiction. So all are in the semi plane $\textbf{Im}(z)\ge0$. Without loss of generality, it's possible to assume $\omega_k$s are sorted counterclockwise around the circle. Meaning the least angle is had by $\omega_1$ and the most angle is had by $\omega_n$. With a clockwise rotation of all of the vectors with the angle of $\omega_1$ problem assumptions still remain and $\omega_1$ becomes equal to $1$. Lemma 4.0 If $u,v$ be two unit vectors that $\lvert u+v\rvert\ge1$ in this case the angle between $u,v$ is at most 120 degrees. It's proved with the cosine theorem. For all $k\ge2$ we have: $$n-1=\lvert\sum\omega_j\rvert\le\lvert\omega_1+\omega_k\rvert+\sum_{j\neq k,1}\lvert\omega_j\rvert=n-2+\lvert\omega_1+\omega_k\rvert\rightarrow1\le\lvert\omega_1+\omega_k\rvert.$$By lemma 3 it's possible to assume that $\omega_1=1$ and also all of $\arg(\omega_k)\in[0,\pi]$. So by recent equation from lemma 4 it concluded that $\arg(\omega_k)\in[0,\frac{2\pi}{3}]$. If $\omega_j$ be that $\arg(\omega_j)\in(\frac{\pi}{3},\frac{2\pi}{3}]$ in this case $\arg(\omega_j^2)\in(\frac{2\pi}{3},\frac{4\pi}{3}]$ So from lemma 4 we have $\lvert\omega_1^2+\omega_j^2\rvert=\lvert1+\omega_j^2\rvert\le1$. So $$n-1\ge\lvert\omega_1^2+\omega_j^2\rvert+\sum_{l\neq j,1}\lvert\omega_l^2\rvert\ge\sum\lvert\omega_l^2\rvert=n-1.$$So equality case of triangle inequality occurs and therefore $$\omega_1=1,\omega_j=\textbf{cis}(\frac{2\pi}{3}),\omega_2=\cdots=\omega_{j-1}=\omega_{j+1}=\cdots=\omega_n=\textbf{cis}(\frac{\pi}{3})$$And with writing the problem condition in this case it's concluded that the only case is $n=2$. So there is no that $j$ and therefore for every index $j$ we have $\arg(\omega_j)\in[0,\frac{\pi}{3}]$ So all the vectors $\omega_k$ lie in a 60-degree range. Now imagine Ali and Abbas starting from the coordinate's origin start to walk respectively with unit steps and in the direction of vectors $\omega_j$ and $\omega_j^2$. The change of angles of Ali is half of the change of angles of Abbas. Through induction and Hinge's theorem in every step, it's possible to conclude in step $k$ Ali is further than Abbas from the origin. (Intuitive movement on a straight line means with the least angle change has the most distance from the origin and however much angle change be more with the condition that the movement is not self-intersecting (Here we used angles are not obtuse) our distance becomes less). So $$\left\lvert\sum\omega_j\right\rvert>\left\lvert\sum\omega_j^2\right\rvert$$that is contradiction.$\blacksquare$