Let $n$ be a natural number. An integer $a>2$ is called $n$-decomposable, if $a^n-2^n$ is divisible by all the numbers of the form $a^d+2^d$, where $d\neq n$ is a natural divisor of $n$. Find all composite $n\in \mathbb{N}$, for which there's an $n$-decomposable number.
To solve this problem, we need to find all composite \( n \in \mathbb{N} \) for which there exists an \( n \)-decomposable number \( a > 2 \). An integer \( a \) is called \( n \)-decomposable if \( a^n - 2^n \) is divisible by all numbers of the form \( a^d + 2^d \), where \( d \neq n \) is a natural divisor of \( n \).

1. **Case: \( n = 2^k \) where \( k > 1 \)**

   For \( n = 2^k \), we can write:
   \[
   a^{2^k} - 2^{2^k} = (a - 2)(a + 2)\left(a^2 + 2^2\right)\left(a^{2^2} + 2^{2^2}\right) \cdots \left(a^{2^{k-1}} + 2^{2^{k-1}}\right).
   \]
   Each term \( a^{2^i} + 2^{2^i} \) for \( i = 0, 1, \ldots, k-1 \) divides \( a^{2^k} - 2^{2^k} \). Therefore, \( n = 2^k \) works.

2. **Case: \( n \) is not a power of two**

   Suppose \( n = 2^k \cdot n' \) where \( n' > 1 \) is odd and \( k \geq 0 \).

   - **Subcase: \( k > 0 \)**

     Consider \( d = 2^k \). We have:
     \[
     a^{2^k} \equiv -2^{2^k} \pmod{a^{2^k} + 2^{2^k}} \implies a^n \equiv -2^n \pmod{a^{2^k} + 2^{2^k}}.
     \]
     This implies:
     \[
     a^{2^k} + 2^{2^k} \mid a^n + 2^n.
     \]
     Since \( a^n \equiv -2^n \pmod{a^{2^k} + 2^{2^k}} \), we get:
     \[
     a^{2^k} + 2^{2^k} \mid 2^{n+1}.
     \]
     Therefore, \( a^{2^k} + 2^{2^k} = 2^u \) for some \( u \). Clearly, \( a \) must be even. Let \( a = 2a' \) where \( a' > 1 \). Then:
     \[
     2^{2^k} \left( (a')^{2^k} + 1 \right) = 2^u \implies (a')^{2^k} + 1 = 2^t \quad \text{for some} \quad t \geq 2.
     \]
     However, since \( k > 0 \) and \( t \geq 2 \), this is impossible modulo 4. Thus, \( n \) cannot be of the form \( 2^k \cdot n' \) with \( n' > 1 \) odd.

   - **Subcase: \( k = 0 \) (i.e., \( n \) is odd)**

     Let \( n = pq \) for some \( p, q > 1 \) odd. We have:
     \[
     a^p \equiv -2^p \pmod{a^p + 2^p} \implies a^n \equiv -2^n \pmod{a^p + 2^p} \implies a^p + 2^p = 2^u \quad \text{for some} \quad u.
     \]
     Again, setting \( a = 2a' \) for some \( a' > 1 \), we find:
     \[
     (a')^p + 1 = 2^t \quad \text{for some} \quad t.
     \]
     Clearly, \( p > 1 \) and \( (a')^{p-1} - (a')^{p-2} + \cdots + 1 \) is a factor of this expression, which is odd and greater than one, yielding a contradiction.

Therefore, the only possible values for \( n \) are of the form \( n = 2^k \) where \( k > 1 \).

The final answer is \( \boxed{ n = 2^k } \) where \( k > 1 \).