Let $D_n$ be the set of divisors of $n$. We need to find all natural numbers $n$ such that it is possible to split $D_n$ into two disjoint sets $A$ and $G$, both containing at least three elements each, where the elements in $A$ form an arithmetic progression and the elements in $G$ form a geometric progression.

We will analyze two main cases:

### Case 1: $1 \in A$
Suppose $A = \{1, 1+k, 1+2k, \ldots, 1+mk\}$ for some integer $k$.

#### Subcase 1a: $n \in A$
If $n \in A$, then $n = 1 + mk$ for some $m$. However, this implies that $n$ and $1 + (m-1)k$ are consecutive terms in the arithmetic progression, which leads to a contradiction because their greatest common divisor must be 1, but $n$ is a multiple of $k$.

#### Subcase 1b: $n \in G$
If $G = \{s, sq, sq^2, \ldots, sq^z = n\}$, then the least common multiple of the elements in $A$ must divide $n$. If $s = 1$, then $G$ contains $1$, which contradicts the disjointness of $A$ and $G$. If $s > 1$, then $q^{z-t}$ for $t < z$ must belong to $A$, but this leads to further contradictions regarding the divisibility and properties of the elements in $A$ and $G$.

### Case 2: $1 \in G$
Suppose $G = \{1, q, q^2, \ldots, q^d\}$.

#### Subcase 2a: $n \in G$
If $n = q^d$, then $q$ must not be prime, as otherwise $A$ and $G$ would not be disjoint. Let $t$ be the smallest divisor of $q$ greater than 1. Then $A$ must contain $t$ and other elements forming an arithmetic progression. However, this leads to contradictions regarding the divisibility and properties of the elements in $A$ and $G$.

#### Subcase 2b: $n \in A$
If $n \in A$, then for some $k$, $n - k \in A$. This implies $n - k$ divides $n$, leading to $n - k \leq k$, which implies $|A| \leq 2$, contradicting the requirement that $A$ must contain at least three elements.

After analyzing all possible cases, we conclude that there are no natural numbers $n$ that satisfy the given conditions.

The answer is: \boxed{\text{No such } n \text{ exists}}.