Problem 1122

For $n=1$, $S_{1}=\{1,2\}$, taking $M_{1}=\{1,2\}$ satisfies the problem's requirements; for $n=2$, $S_{2}=\{1,2,3,4,5\}$, taking $M_{2}=\{1, 2, 4, 5\}$ satisfies the problem's requirements; for $n=3$, $S_{3}=\{1,2,3, \cdots, 14\}$, taking $M_{3}=\{1,2,4,5,10,11,13,14\}$ satisfies the problem's requirements, and it is noted that if $x, y, z$ do not form an arithmetic sequence, then for any $a$, $a, a+x, a+y, a+z$ also do not form an arithmetic sequence, and the elements of $M_{3}$ have the following relationships: $10=1+3^{2}, 11=2+3^{2}, 13=4+3^{2}, 14=5+3^{2}$.

Below, we use an inductive construction method to prove that a subset $M_{n}$ satisfying the conditions exists.

For $n=1$, as previously stated, $M_{1}$ satisfies the conditions.
Assume for a positive integer $n$, there exists a subset $M_{n}$ of $S_{n}$ containing $2^{n}$ elements such that no three elements in $M_{n}$ form an arithmetic sequence. Let $M_{n+1}=M_{n} \cup\left\{3^{n}+a \mid a \in M_{n}\right\}$.
Then $M_{n+1}$ contains $2\left|M_{n}\right|=2 \cdot 2^{n}=2^{n+1}$ elements, and the largest element in $M_{n+1}$ does not exceed $\frac{1}{2}\left(3^{n}+1\right) + 3^{n} = \frac{1}{2}\left(3^{n+1}+1\right)$, so $M_{n+1}$ is a subset of $S_{n+1}$ containing $3^{n+1}$ elements. If there exist three numbers $x, y, z (x<y<z)$ in $M_{n+1}$ that form an arithmetic sequence, then $x, y, z$ cannot all belong to $M_{n}$, nor can they all belong to $\left\{3^{n}+a \mid a \in M_{n}\right\}$, hence only $x \in M_{n}, z \in\left\{3^{n}+a \mid a \in M_{n}\right\}$.

If $y \in M_{n}$, then by $y \leqslant \frac{1}{2}\left(3^{n}+1\right), x \geqslant 1$, we get
$$2=2 y-x \leqslant 2 \times \frac{1}{2}\left(3^{n}+1\right)-1=3^{n},$$

which contradicts $z \in\left\{3^{n}+a \mid a \in M_{n}\right\}, z \geqslant 3^{n}+1$.
If $y \in\left\{3^{n}+a \mid a \in M_{n}\right\}$, then by $y \geqslant 3^{n}+1, z \leqslant \frac{1}{2}\left(3^{n+1}+1\right)$ we get
$$x=2 y-z \geqslant 2\left(3^{n}+1\right)-\frac{1}{2}\left(3^{n+1}+1\right)=\frac{1}{2}\left(3^{n}+3\right),$$

which contradicts $x \in M_{n}, x \leqslant \frac{1}{2}\left(3^{n}+1\right)$.
This proves that no three elements in $M_{n+1}$ form an arithmetic sequence. Thus, we have used an inductive construction method to prove that a subset $M_{n}$ satisfying the problem's conditions exists.