Find all natural $n{}$ such that for every natural $a{}$ that is mutually prime with $n{}$, the number $a^n - 1$ is divisible by $2n^2$.
1. **Prove that \( n \) is even:**
   - Assume \( n \) is odd.
   - Take \( a \equiv 1 \pmod{n} \) and \( a \equiv 0 \pmod{2} \). Such an \( a \) exists via the Chinese Remainder Theorem (CRT).
   - Clearly, \( 2 \nmid a^n - 1 \) for such an \( n \), so \( n \) must be even.

2. **Prove that \( n \) is square-free:**
   - Let \( p \) be a prime dividing \( n \). We need to show that \( v_p(n) = 1 \).
   - First, consider \( p > 2 \). Take \( a \equiv p + 1 \pmod{p^2} \) and \( a \equiv 1 \pmod{n/p^e} \), where \( e = v_p(n) \).
   - By the Lifting The Exponent (LTE) lemma, \( v_p(a^n - 1) = 1 + v_p(n) \).
   - Since \( v_p(2n^2) = 2v_p(n) \), we get \( v_p(n) = 1 \).

3. **Prove that \( 2 \mid \mid n \):**
   - Note that \( v_2(a^n - 1) = v_2(a - 1) + v_2(a + 1) + v_2(n) - 1 \) via LTE.
   - We must ensure \( v_2(a - 1) + v_2(a + 1) \ge v_2(n) + 2 \) for any such \( a \).
   - Take \( a \equiv 1 \pmod{n'} \) and \( a \equiv 3 \pmod{8} \), where \( n' \) is the odd part of \( n \). This immediately leads to a contradiction.
   - Therefore, \( n \) is of the form \( 2p_1 \cdots p_k \) where \( 2 < p_1 < \cdots < p_k \) are primes.

4. **Determine the specific form of \( n \):**
   - Take \( n \) such that \( n \equiv 1 \pmod{n/p_i} \) and \( n \equiv g_i \pmod{p_i} \) where \( g_i \) is a primitive root modulo \( p_i \).
   - This implies \( p_i - 1 \mid 2p_1 \cdots p_k \).
   - Noting that \( (p_i - 1, p_{i+1} \cdots p_k) = 1 \), we conclude \( p_i - 1 \mid 2p_1 \cdots p_{i-1} \) for any \( i \).

5. **Reverse-engineer the solutions:**
   - From the above conditions, we can deduce the specific values of \( n \).
   - The solutions are \( n = 2, 2 \cdot 3, 2 \cdot 3 \cdot 7, 2 \cdot 3 \cdot 7 \cdot 43 \).

The final answer is \( \boxed{2, 6, 42, 1806} \).