Four, let $x, y$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}$, $\cdots, a_{n}$, and satisfy the condition: when $x>y$, $x-y \geqslant \frac{x}{15}$. Find the maximum number of these natural numbers $n$.
$$
\begin{array}{l}
\text { Let's assume } a_{1}=1, d_{4} \geqslant 2 ; \\
a_{5}=a_{4}+d_{4} \geqslant 6, d_{5} \geqslant \frac{6^{2}}{19-6}>2, d_{5} \geqslant 3 ; \\
a_{6} \geqslant 9, d_{6} \geqslant \frac{9^{2}}{19-9}>8, d_{6} \geqslant 9 ; \\
a_{7} \geqslant 18, d_{7} \geqslant \frac{18^{2}}{19-18}=324 ; \\
a_{8} \geqslant 342>19 .
\end{array}
$$

From this, we can see that $n-1 \leqslant 7, n \leqslant 8$.
On the other hand, the 8 natural numbers $1,2,3,4,6,9,18,342$ meet the conditions of the problem.
Therefore, the maximum value of $n$ is 8.