Three, (25 points) Let $x, y (x>y)$ be any two numbers in a set of distinct natural numbers $a_{1}, a_{2}, \cdots, a_{n}$, satisfying $x-y \geqslant \frac{xy}{31}$. Find the maximum value of the number of elements $n$ in this set of natural numbers.
Let's assume $a_{1}1 \\
\Rightarrow d_{6} \geqslant 2 \Rightarrow a_{7}=a_{6}+d_{6} \geqslant 8 \\
\Rightarrow d_{7} \geqslant \frac{8^{2}}{31-8}>2 \Rightarrow d_{7} \geqslant 3 . \\
\Rightarrow a_{8}=a_{7}+d_{7} \geqslant 11 \\
\Rightarrow d_{8} \geqslant \frac{11^{2}}{31-11}>6 \Rightarrow d_{8} \geqslant 7 \\
\Rightarrow a_{9}=a_{8}+d_{8} \geqslant 18 \\
\Rightarrow d_{9} \geqslant \frac{18^{2}}{31-18}>24 \Rightarrow d_{9} \geqslant 25 \\
\Rightarrow a_{10}=a_{9}+d_{9} \geqslant 43>31 .
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Therefore, $n-1 \leqslant 9 \Rightarrow n \leqslant 10$.
At the same time, the ten natural numbers $1,2,3,4,5,6,8, 11,18, 43$ meet the conditions of the problem.
Thus, the maximum value of $n$ is 10.