Prove that for natural number $n$ it's possible to find complex numbers $\omega_1,\omega_2,\cdots,\omega_n$ on the unit circle that
$$\left\lvert\sum_{j=1}^{n}\omega_j\right\rvert=\left\lvert\sum_{j=1}^{n}\omega_j^2\right\rvert=n-1$$
iff $n=2$ occurs.
1. **Lemma:**
   For any \( n \) complex numbers on the unit circle, we have:
   \[
   \sum_{1 \leq j < k \leq n} \lvert \omega_j - \omega_k \rvert^2 = n^2 - \left\lvert \sum_{j=1}^{n} \omega_j \right\rvert^2
   \]
   **Proof:**
   \[
   \sum_{1 \leq j < k \leq n} \lvert \omega_j - \omega_k \rvert^2 = \sum_{1 \leq j < k \leq n} (\omega_j - \omega_k) \overline{(\omega_j - \omega_k)}
   \]
   \[
   = (n-1) \sum_{j=1}^{n} \omega_j \overline{\omega_j} + \sum_{1 \leq j < k \leq n} (\omega_j \overline{\omega_k} + \overline{\omega_j} \omega_k)
   \]
   Since \(\omega_j\) are on the unit circle, \(\omega_j \overline{\omega_j} = 1\). Therefore,
   \[
   (n-1) \sum_{j=1}^{n} \omega_j \overline{\omega_j} = (n-1)n
   \]
   Also,
   \[
   \left\lvert \sum_{j=1}^{n} \omega_j \right\rvert^2 = \left( \sum_{j=1}^{n} \omega_j \right) \left( \sum_{j=1}^{n} \overline{\omega_j} \right) = n + \sum_{1 \leq j < k \leq n} (\omega_j \overline{\omega_k} + \overline{\omega_j} \omega_k)
   \]
   Subtracting the two preceding equations, we get:
   \[
   \sum_{1 \leq j < k \leq n} \lvert \omega_j - \omega_k \rvert^2 = n^2 - \left\lvert \sum_{j=1}^{n} \omega_j \right\rvert^2
   \]
   This proves the lemma.

2. **Assume that \(\omega_1, \omega_2, \cdots, \omega_n\) exist with the problem's condition.**
   The lemma implies that:
   \[
   \sum_{1 \leq j < k \leq n} \lvert \omega_j - \omega_k \rvert^2 = \sum_{1 \leq j < k \leq n} \lvert \omega_j^2 - \omega_k^2 \rvert^2 = 2n - 1
   \]
   \[
   \implies \sum_{1 \leq j < k \leq n} \lvert \omega_j - \omega_k \rvert^2 \lvert \omega_j + \omega_k \rvert^2 = \sum_{1 \leq j < k \leq n} \lvert \omega_j - \omega_k \rvert^2
   \]

3. **Assume \(a, b\) such that \(\lvert \omega_a + \omega_b \rvert\) is minimum.**
   Because all of the \(\omega_j\)'s are not equal, we have \(\lvert \omega_a + \omega_b \rvert \leq 1\).
   And we have:
   \[
   n-1 = \left\lvert \sum_{j=1}^{n} \omega_j \right\rvert \leq \lvert \omega_a + \omega_b \rvert + \sum_{j \not\in \{a, b\}} \lvert \omega_j \rvert = n-2 + \lvert \omega_a + \omega_b \rvert
   \]
   \[
   \implies \lvert \omega_a + \omega_b \rvert \geq 1
   \]
   So the equality case occurs and \(\{\omega_j\}_{j \not\in \{a, b\}}\) are on the internal angle bisector of \(\omega_a\), \(\omega_b\).

4. **Observation of \(\omega_j^2\):**
   We see that all \(\{\omega_j^2\}_{j \not\in \{a, b\}}\) are on the external angle bisector of \(\omega_a^2\), \(\omega_b^2\).

5. **Equality case:**
   \[
   n-1 = \left\lvert \sum_{j=1}^{n} \omega_j^2 \right\rvert \leq \lvert \omega_a^2 + \omega_b^2 \rvert + \sum_{j \not\in \{a, b\}} \lvert \omega_j^2 \rvert = n-1
   \]
   So again the equality holds and \(\{\omega_j^2\}_{j \not\in \{a, b\}}\) should be on the internal angle bisector of \(\omega_a^2\), \(\omega_b^2\). Therefore, \(\{1, 2, \ldots, n\} \setminus \{a, b\} = \emptyset \implies n = 2\).

6. **Conclusion:**
   The equality case is when \(n = 2\) and \(\frac{\omega_a}{\omega_b} \in \{\text{cis}(\frac{2\pi}{3}), \text{cis}(\frac{-2\pi}{3})\}\).

The final answer is \( \boxed{ n = 2 } \).