3. (China) Let $S=\{1,2,3, \ldots, 280\}$. Find the smallest natural number $n$ such that every subset of $S$ with $n$ elements contains 5 pairwise coprime numbers.
Let $A_{1}=\{S \mapsto$ all natural numbers divisible by $i\}, i=2,3,5,7$. Let $A=A_{2} \cup A_{3} \cup A \cup \cup A_{7}$. Using the principle of inclusion-exclusion, it is easy to calculate that the number of elements in $A$ is 216. Since any 5 numbers chosen from $A$ must have at least two numbers in the same $A_i$, they are not coprime, thus $n \geqslant 217$.
$$
\begin{array}{l} 
\text { On the other hand, let } \\
B_{1}=\{1 \text { and all prime numbers in } S\}, \\
B_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\}, \\
B_{3}=\{2 \times 131,3 \times 89,5 \times 53, \\
7 \times 37,11 \times 23,13 \times 19\}, \\
B_{4}=\{2 \times 127,3 \times 83,5 \times 47, \\
7 \times 31,11 \times 19,13 \times 17\}, \\
B_{5}=\{2 \times 113,3 \times 79,5 \times 43, \\
7 \times 29,11 \times 17\} . \\
B_{8}=\{2 \times 109,3 \times 73,5 \times 41, \\
7 \times 23,11 \times 13\} .
\end{array}
$$

It is easy to see that the number of elements in $B_{1}$ is 60. Let $B=B_{1} \cup B_{2} \cup B_{3} \cup B_{4} \cup B_{5} \cup B_{8}$, then the number of elements in $B$ is 88. Thus, the number of elements in $S-B$ is 192. When 217 numbers are chosen from $S$, since $217-192=25$, there exists $1 \leqslant i \leqslant 6$ such that these 217 numbers include 5 numbers from $B$, and these 5 numbers are pairwise coprime. Therefore, $n \leqslant 217$. Hence, we have $n=217$.