Five, color the numbers in $S=\{0,1,2, \cdots, n\}$ with two colors arbitrarily. Find the smallest positive integer $n$, such that there must exist $x, y, z \in S$ of the same color, satisfying $x+y=2 z$.
Let $A=\{0,2,5,7\}$ and $B=\{1,3,4,6\}$. Then neither $A$ nor $B$ contains a 3-term arithmetic progression. By coloring the elements of $A$ red and the elements of $B$ yellow, we can see that $n>7$.

When $n \geqslant 8$, we will prove that there must be a color for which the numbers form a 3-term arithmetic progression.

In fact, we only need to consider the first 9 numbers $0,1, \cdots, 8$. Suppose these numbers are divided into two subsets $A$ and $B$ (i.e., they are colored with two different colors), and without loss of generality, assume $0 \in A$.
Assume the conclusion is false.
(1) Suppose $1 \in A$, then $2 \in B$.
(i) If $3 \in A$, then by $5+1=2 \times 3$, we have $5 \in B$. Similarly, $\Rightarrow$ $8 \in A \Rightarrow 4 \in B \Rightarrow 6 \in A$, and by $0, 3, 6 \in A$ we reach a contradiction.
(ii) If $3 \in B$, then by $2+4=2 \times 3$, we have $4 \in A$. Similarly, $\Rightarrow$ $7 \in B \Rightarrow 5 \in A \Rightarrow 6 \in B \Rightarrow 8 \in A$, and by $0, 4, 8 \in A$ we reach a contradiction.
(2) Suppose $1 \in B$.
(i) If $2 \in B$, then by $1+3=2 \times 2$, we have $3 \in A$. Similarly, $\Rightarrow$ $6 \in B \Rightarrow 4 \in A \Rightarrow 5 \in B, 8 \in B$, and by $2, 5, 8 \in B$ we reach a contradiction.
(ii) If $2 \in A$, then by $0+4=2 \times 2$, we have $4 \in B$. Similarly, $\Rightarrow$ $7 \in A$.

If $3 \in B$, then $5 \in A \Rightarrow 6 \in B \Rightarrow 8 \in A$, and by $2, 5, 8 \in A$ we reach a contradiction.

If $3 \in A$, then $5 \in B \Rightarrow 6 \in A$, and by $0, 3, 6 \in A$ we reach a contradiction.
Therefore, $n_{\min }=8$.