2・102 Let the set $S=\{1,2, \cdots, 50\}$, find the smallest positive integer $n$, such that in every $n$-element subset of $S$ there are 3 numbers that can serve as the 3 side lengths of a right triangle.
[Solution] First, we prove the following lemma.

Lemma If positive integers $x, y, z$ satisfy the equation
$$
x^{2}+y^{2}=z^{2} \text {, }
$$

then at least one of the three numbers is a multiple of 5.
Proof of the lemma: Because
$$
\begin{array}{l}
(5 k+1)^{2}=25 k^{2}+10 k+1 \equiv 1, \\
(5 k+2)^{2}=25 k^{2}+20 k+4 \equiv-1, \\
(5 k+3)^{2}=25 k^{2}+30 k+9 \equiv-1, \\
(5 k+4)^{2}=25 k^{2}+40 k+16 \equiv 1,
\end{array}
$$

if $x$ and $y$ are not multiples of 5, then $x^{2}$ and $y^{2}$ are congruent to 1 or -1 modulo 5. Thus, $z^{2}$ can only be congruent to 0 modulo 5, so $z$ is a multiple of 5.

Returning to the solution of the original problem. Consider all Pythagorean triples where 10, 15, 25, 40, and 45 are one of the sides of a right triangle. Since the positive integer solutions to equation (1) 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 $k, a, b \in \mathbb{N}$, and $(a, b)=1, a>b$, we know that there are the following 11 Pythagorean triples:
$$
\begin{array}{l}
(10,8,6),(26,24,10),(15,12,9),(17,15,8), \\
(39,36,15),(25,24,7),(40,32,24),(41,40,9), \\
(45,36,27),(25,20,15),(50,40,30) .
\end{array}
$$

Notice that in the first 9 Pythagorean triples, each group contains one of the numbers 8, 9, 24, or 36. Therefore, the set
$$
M=S-\{5,8,9,20,24,30,35,36,50\}
$$

contains no Pythagorean triple of any 3 numbers. Thus, the smallest positive integer $n \geqslant 42$.
On the other hand, in the following 9 Pythagorean triples
$$
\begin{array}{l}
(3,4,5),(7,24,25),(8,15,17),(9,40,41), \\
(12,35,37),(14,48,50),(16,30,34),(20,21,29), \\
(27,36,45)
\end{array}
$$

the 27 numbers are all distinct. Therefore, for any 42-element subset $M$ of $S$, the 8 numbers not in $M$ can belong to at most 8 of the 9 groups in (3), so at least one Pythagorean triple is entirely in $M$.
In conclusion, the smallest positive integer $n=42$.