Let $n$ be a fixed natural number.
Find all $n$ tuples of natural pairwise distinct and coprime numbers like $a_1,a_2,\ldots,a_n$ such that for $1\leq i\leq n$ we have
\[ a_1+a_2+\ldots+a_n|a_1^i+a_2^i+\ldots+a_n^i  \]
1. **Consider the polynomial \( P(t) \):**
   \[
   P(t) = t(t - a_1)(t - a_2) \cdots (t - a_n) = t^{n+1} + c_n t^n + c_{n-1} t^{n-1} + \cdots + c_0 t
   \]
   This polynomial has roots \( a_1, a_2, \ldots, a_n \).

2. **Evaluate \( P(t) \) at \( t = a_i \):**
   \[
   P(a_i) = 0 \quad \text{for} \quad i = 1, 2, \ldots, n
   \]
   Therefore,
   \[
   0 = P(a_1) + P(a_2) + \cdots + P(a_n)
   \]
   Expanding this, we get:
   \[
   0 = a_1^{n+1} + a_2^{n+1} + \cdots + a_n^{n+1} + c_n (a_1^n + a_2^n + \cdots + a_n^n) + \cdots + c_0 (a_1 + a_2 + \cdots + a_n)
   \]

3. **Given condition:**
   \[
   a_1 + a_2 + \cdots + a_n \mid a_1^i + a_2^i + \cdots + a_n^i \quad \text{for} \quad 1 \leq i \leq n
   \]
   This implies:
   \[
   a_1 + a_2 + \cdots + a_n \mid c_n (a_1^n + a_2^n + \cdots + a_n^n) + \cdots + c_0 (a_1 + a_2 + \cdots + a_n)
   \]
   Hence,
   \[
   a_1 + a_2 + \cdots + a_n \mid a_1^{n+1} + a_2^{n+1} + \cdots + a_n^{n+1}
   \]

4. **Inductive proof:**
   Using the polynomial \( P(t) = t^{k-n}(t - a_1)(t - a_2) \cdots (t - a_n) \), we can inductively prove that:
   \[
   a_1 + a_2 + \cdots + a_n \mid a_1^k + a_2^k + \cdots + a_n^k \quad \text{for} \quad k \in \mathbb{Z}_+
   \]

5. **Divisibility by \( n(n-1) \):**
   Let \( S = a_1 + a_2 + \cdots + a_n \). We need to show that \( S \mid n(n-1) \).

6. **Prime factorization:**
   Let \( S = p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m} \). We show that \( p_i^{\alpha_i} \mid n \) or \( p_i^{\alpha_i} \mid n-1 \) for \( i = 1, 2, \ldots, m \).

7. **Case 1: \( a_j \) not divisible by \( p_i \):**
   \[
   a_j^{\phi(p_i^{\alpha_i})} \equiv 1 \pmod{p_i^{\alpha_i}}
   \]
   Plugging \( k = \phi(p_i^{\alpha_i}) \) in the inductive result:
   \[
   0 \equiv a_1 + a_2 + \cdots + a_n \equiv n \pmod{p_i^{\alpha_i}}
   \]
   Thus, \( n \) is divisible by \( p_i^{\alpha_i} \).

8. **Case 2: \( a_j \) divisible by \( p_i \):**
   \[
   a_j^{\phi(p_i^{\alpha_i})} \equiv 0 \pmod{p_i^{\alpha_i}}
   \]
   Since \( a_1, a_2, \ldots, a_n \) are pairwise coprime, no other \( a_k \) is divisible by \( p_i \). Thus:
   \[
   0 \equiv a_1 + a_2 + \cdots + a_n \equiv n-1 \pmod{p_i^{\alpha_i}}
   \]
   Hence, \( n-1 \) is divisible by \( p_i^{\alpha_i} \).

9. **Conclusion:**
   \[
   S \mid n(n-1)
   \]

10. **Contradiction:**
    If \( S \neq n(n-1) \), then:
    \[
    S \leq \frac{n(n-1)}{2}
    \]
    But since \( a_1, a_2, \ldots, a_n \) are distinct:
    \[
    S \geq 1 + 2 + \cdots + n = \frac{n(n+1)}{2}
    \]
    This leads to a contradiction.

11. **Special cases:**
    For \( n \geq 2 \), we check specific values and find contradictions. For \( n = 1 \), \( a_1 \) can be any natural number.

The final answer is there are no solutions if \( \boxed{ n \geq 2 } \) and if \( n = 1 \), \( a_1 \) is arbitrary.