Find all natural numbers $n\ge1$ such that the implication
$$(11\mid a^n+b^n)\implies(11\mid a\wedge11\mid b)$$holds for any two natural numbers $a$ and $b$.
To solve the problem, we need to find all natural numbers \( n \ge 1 \) such that the implication 
\[ (11 \mid a^n + b^n) \implies (11 \mid a \wedge 11 \mid b) \]
holds for any two natural numbers \( a \) and \( b \).

1. **Case 1: \( n \) is odd**

   If \( n \) is odd, we can use the fact that \( a + b \mid a^n + b^n \). Specifically, if \( a \equiv -b \pmod{11} \), then \( a + b \equiv 0 \pmod{11} \), and thus \( 11 \mid a + b \). This implies \( 11 \mid a^n + b^n \).

   For example, let \( a \equiv 1 \pmod{11} \) and \( b \equiv -1 \pmod{11} \). Then:
   \[
   a^n + b^n \equiv 1^n + (-1)^n \equiv 1 - 1 \equiv 0 \pmod{11}
   \]
   Hence, \( 11 \mid a^n + b^n \), but \( 11 \nmid a \) and \( 11 \nmid b \). Therefore, the implication fails for odd \( n \).

2. **Case 2: \( n \) is even**

   Assume \( n \) is even. Let \( n = 2k \) for some integer \( k \). Define \( u = a^k \) and \( v = b^k \). Then:
   \[
   a^n + b^n = (a^k)^2 + (b^k)^2 = u^2 + v^2
   \]
   Given \( 11 \mid a^n + b^n \), we have \( 11 \mid u^2 + v^2 \).

   Suppose \( 11 \nmid u \) and \( 11 \nmid v \). Then \( u \) and \( v \) are invertible modulo 11. Let \( w = u \cdot v^{-1} \). Then:
   \[
   (u \cdot v^{-1})^2 \equiv -1 \pmod{11}
   \]
   This implies that there exists an integer \( w \) such that \( w^2 \equiv -1 \pmod{11} \). However, 11 is congruent to 3 modulo 4, and by Fermat's theorem on sums of two squares, \(-1\) is not a quadratic residue modulo a prime \( p \equiv 3 \pmod{4} \). Therefore, \( w^2 \equiv -1 \pmod{11} \) has no solution, leading to a contradiction.

   Hence, if \( 11 \mid u^2 + v^2 \), then either \( 11 \mid u \) or \( 11 \mid v \). Since \( u = a^k \) and \( v = b^k \), it follows that \( 11 \mid a \) and \( 11 \mid b \).

Therefore, the implication holds for all even \( n \).

The final answer is all even \( \boxed{ n } \).