Example 6 Let the set $S=\{1,2, \cdots, 50\}, X$ be any subset of $S$, $|X|=n$. Find the smallest positive integer $n$, such that the set $X$ must contain three numbers that are the lengths of the three sides of a right-angled triangle.
Let the three sides of a right-angled triangle be $x, y, z$, with $x^{2}+y^{2}=z^{2}$,
the positive integer solutions can be expressed as
$$
x=k\left(a^{2}-b^{2}\right), y=2 k a b, z=k\left(a^{2}+b^{2}\right)
$$

where $k, a, b \in \mathbf{N}^{*}$ and $(a, b)=1, a>b$.
Among $x, y, z$, there must be one that is a multiple of 5.
Otherwise, if $a, b, c$ are not multiples of 5, then $a, b, c$ are numbers of the form $5 m \pm 1, 5 m \pm 2$ $(m \in \mathbf{N})$, then $a^{2} \equiv \pm 1(\bmod 5), b^{2} \equiv \pm 1(\bmod 5), c^{2} \equiv \pm 1(\bmod 5)$, and $c^{2}=a^{2}+b^{2} \equiv 0$ or $\pm 2$, which is a contradiction!
Let the set $A=\{S$ of all numbers coprime with 5 $\}$, then $\operatorname{Card} A=40$.
If $10,15,25,40,45$ are used as the lengths of one side of a right-angled triangle, then by (1) we can find the corresponding sides in $A$ to form the following right-angled triangles:
$$
(10,8,6),(26,24,10),(15,12,9),(17,15,8),(39,36,15),(25,24,7),(40,32,24), (41,40,9), (42,27,36).
$$
In addition, there are no numbers in $A$ that can form the three sides of a right-angled triangle with $10,15,25,40,45$.
Let $M=A \cup\{10,15,25,40,45\} /\{8,9,24,36\}$, then $\operatorname{Card} M=41$.
From the above, we know that no three numbers in $A$ can form the sides of a right-angled triangle.
Since $M$ does not contain $8,9,24,36$, $10,15,25,40,45$ cannot find numbers in $M$ to form the three sides of a right-angled triangle, i.e., any three numbers in $M$ cannot form the sides of a right-angled triangle. Therefore, $n \geqslant 42$.
Furthermore, by the integer solutions in (1), we can form the set
$$
B=\{\underline{3,4,5}, \underline{17,15,8}, \underline{29,21,20}, \underline{25,24,7}, \underline{34,16,30}, \underline{37,35,12}, \underline{50,48,14}, \underline{41,40,9},
$$
$\underline{45,36,27\},}$
where the three numbers underlined can form the sides of a right-angled triangle, $\operatorname{Card} B=27$.
The number of elements in $S / B$ is $50-27=23$. If 42 numbers are chosen from $S$, since $42-23=19$, the 42 chosen numbers must contain 19 numbers from $B$, so at least one set of three numbers underlined in $B$ will be among the 42 chosen numbers, i.e., any 42 chosen numbers will contain at least three numbers that can form the sides of a right-angled triangle. Therefore, the minimum value of $n$ is 42.