Find all natural numbers $n$ for which $1^{\phi (n)} + 2^{\phi (n)} +... + n^{\phi (n)}$ is coprime with $n$.
1. **Understanding the Problem:**
   We need to find all natural numbers \( n \) such that the sum \( 1^{\phi(n)} + 2^{\phi(n)} + \cdots + n^{\phi(n)} \) is coprime with \( n \). Here, \(\phi(n)\) is Euler's totient function, which counts the number of integers up to \( n \) that are coprime with \( n \).

2. **Initial Claim:**
   We claim that the answer is \( n = p_1 p_2 \cdots p_t \), where \( p_i \) are distinct primes. This means \( n \) is a product of distinct prime numbers.

3. **Using Fermat's Little Theorem:**
   Fermat's Little Theorem states that for any integer \( a \) and a prime \( p \), \( a^{p-1} \equiv 1 \pmod{p} \) if \( p \nmid a \). For \( n = p_1 p_2 \cdots p_t \), \(\phi(n) = (p_1 - 1)(p_2 - 1) \cdots (p_t - 1)\).

4. **Sum Modulo \( p \):**
   Consider the sum \( A = \sum_{i=1}^{n} i^{\phi(n)} \). We need to check if \( \gcd(A, n) = 1 \). If \( \gcd(A, n) \neq 1 \), then there exists a prime \( p \) such that \( p \mid A \) and \( p \mid n \).

5. **Applying Fermat's Little Theorem:**
   For each prime \( p_i \mid n \), we have:
   \[
   i^{\phi(n)} \equiv i^{(p_1-1)(p_2-1) \cdots (p_t-1)} \pmod{p_i}
   \]
   Since \( \phi(n) \) is a multiple of \( p_i - 1 \), by Fermat's Little Theorem, \( i^{\phi(n)} \equiv 1 \pmod{p_i} \) for \( i \) not divisible by \( p_i \).

6. **Sum Modulo \( p_i \):**
   \[
   A \equiv \sum_{i=1}^{n} i^{\phi(n)} \equiv \sum_{i=1}^{n} 1 \equiv n \pmod{p_i}
   \]
   Since \( n \equiv 0 \pmod{p_i} \), we have \( A \equiv 0 \pmod{p_i} \).

7. **Contradiction for \( p^2 \mid n \):**
   If \( n \) contains a square of a prime, say \( n = y \cdot p^2 \), then:
   \[
   A \equiv y \cdot p^2 - py \equiv 0 \pmod{p}
   \]
   This implies \( \gcd(A, n) \geq p \), which contradicts our requirement that \( \gcd(A, n) = 1 \).

8. **Conclusion:**
   Therefore, \( n \) must be a product of distinct primes, i.e., \( n = p_1 p_2 \cdots p_t \).

The final answer is \( \boxed{ n = p_1 p_2 \cdots p_t } \), where \( p_i \) are distinct primes.