1. Define the sequence and sets:
Let $a_i$ denote the terms in the sequence, with $a_1 = 1$ and $a_2 = 24$. Define the set $S_n := \{a_1, \ldots, a_n\}$ as the set of the first $n$ terms of the sequence, and $S_\infty := \bigcup_{n \in \mathbb{Z^+}} S_n$ as the set of all terms that eventually appear in the sequence. Let $U := \mathbb{Z^+} \setminus S_\infty$ be the set of positive integers that never appear in the sequence.

2. **Assume $U \neq \emptyset$:**
Suppose not all positive integers occur in the sequence, i.e., $U \neq \emptyset$.

3. **Composite numbers in $U$:**
- Assume $u \in U$ is a composite number. By the Well-Ordering Principle (WOP), we can assume $u$ is the smallest composite number in $U$.
- Consider the set $\Psi(u) = \{ \text{numbers } < u \text{ having a common factor with } u \}$.
- Since $u$ is composite, it has at least one prime factor. Therefore, there exists some $v = a_j \in \Psi(u)$ with $j$ maximal (the last number having a common factor with $u$ to appear in the sequence).
- By the sequence's rule, the next term $a_{j+1}$ should be the smallest positive integer not yet in the sequence and not coprime with $a_j$. Since $v$ and $u$ share a common factor, $u$ should be $a_{j+1}$, implying $u \in S_\infty$, which is a contradiction. Therefore, $U$ cannot contain any composite number.

4. **Prime numbers in $U$:**
- Let $p$ be the smallest prime number in $U$ (using WOP again). Note that $p \neq 2$ because $a_3 = 2$.
- Since $2p$ is a composite number, and we have already shown that $U$ cannot contain any composite numbers, $2p \in S_\infty$. Thus, there exists some $j > 3$ such that $2p = a_j$.
- By the sequence's rule, the next term $a_{j+1}$ should be the smallest positive integer not yet in the sequence and not coprime with $a_j$. Since $2p$ and $p$ share a common factor, $p$ should be $a_{j+1}$, implying $p \in S_\infty$, which is a contradiction. Therefore, $U$ cannot contain any prime number.

5. Conclusion:
- Since $U$ cannot contain any composite numbers or prime numbers, and $1 \notin U$, we conclude that $U = \emptyset$.
- Therefore, $S_\infty = \mathbb{Z^+}$, meaning all positive integers occur in the sequence.

$\blacksquare$