Let $A$ be an infinite set of positive integers. Find all natural numbers $n$ such that for each $a \in A$,
\[a^n + a^{n-1} + \cdots + a^1 + 1 \mid a^{n!} + a^{(n-1)!} + \cdots   + a^{1!} + 1.\]

[i]Proposed by Milos Milosavljevic[/i]
1. **Understanding the Problem:**
   We need to find all natural numbers \( n \) such that for each \( a \in A \), the polynomial \( a^n + a^{n-1} + \cdots + a + 1 \) divides \( a^{n!} + a^{(n-1)!} + \cdots + a^{1!} + 1 \).

2. **Expressing the Polynomials:**
   Let \( P(x) = x^n + x^{n-1} + \cdots + x + 1 \). This can be rewritten using the formula for the sum of a geometric series:
   \[
   P(x) = \frac{x^{n+1} - 1}{x - 1}
   \]
   Let \( Q(x) = x^{n!} + x^{(n-1)!} + \cdots + x^{1!} + 1 \).

3. **Divisibility Condition:**
   We need \( P(a) \mid Q(a) \) for all \( a \in A \). Since \( A \) is an infinite set of positive integers, by a well-known lemma in algebra, if \( P(x) \mid Q(x) \) for infinitely many \( x \), then \( P(x) \mid Q(x) \) in \( \mathbb{Z}[x] \).

4. **Roots of Unity:**
   The roots of \( P(x) \) are the \((n+1)\)-th roots of unity, excluding 1. Let \( \omega \) be a primitive \((n+1)\)-th root of unity. Then \( P(\omega) = 0 \).

5. **Evaluating \( Q(x) \) at Roots of Unity:**
   Since \( P(x) \mid Q(x) \), \( Q(\omega) = 0 \). We need to check if \( Q(\omega) = 0 \) for all \((n+1)\)-th roots of unity \( \omega \).

6. **Roots of Unity Filter:**
   Using the roots of unity filter, we can write:
   \[
   \frac{\sum_{i=0}^{n} Q(\omega^i)}{n+1} = 1
   \]
   This implies that the sum of the coefficients of \( Q(x) \) that are divisible by \( n+1 \) must be 1. However, none of \( n!, (n-1)!, \ldots, 1! \) are divisible by \( n+1 \) unless \( n+1 \) is a prime number.

7. **Checking \( n+1 \) as a Prime:**
   Let \( p = n+1 \). Then \( p \) must be a prime number. We need to check if \( Q(\omega) = 0 \) for \( \omega \) being a primitive \( p \)-th root of unity.

8. **Residue System Modulo \( p \):**
   The set \( \{n!, (n-1)!, \ldots, 1!, 0\} \) forms a complete residue system modulo \( p \). We have:
   \[
   n! \equiv -1 \pmod{p}
   \]
   and
   \[
   (n-1)! \equiv 1 \pmod{p}
   \]
   This leads to a contradiction since \( 1! \equiv 1 \pmod{p} \).

9. **Conclusion:**
   The only possible values for \( n \) are 1 and 2. Checking these values, we find that they satisfy the given condition.

\(\blacksquare\)

The final answer is \( \boxed{ n = 1 } \) or \( n = 2 \).