5. Let the set $S=\{1,2, \cdots, 280\}$, find the smallest positive integer $n$, such that any $n$-element subset $T$ of $S$ contains 5 numbers that are pairwise coprime.
5. Let $M_{i}=\{x \mid x \in S$ and $i \mid x\}\left(i \in \mathbf{N}^{*}\right)$.

Let the set $T=M_{2} \cup M_{3} \cup M_{5} \cup M_{7}$, we know that $T$ does not meet the requirement, and $|T|=216$, so the smallest $n \geqslant 217$. Below, we construct a proof that $n \leqslant 217$.
Using the sieve method, we know that there are 59 prime numbers in $S$.
Construct the following drawers:
$$
\begin{array}{l}
A_{1}=\{ \text{all prime numbers in } S \} \cup \{1\} ; \\
A_{2}=\left\{2^{2}, 3^{2}, 5^{2}, 7^{2}, 11^{2}, 13^{2}\right\} ; \\
A_{3}=\{2 \times 41,3 \times 37,5 \times 31,7 \times 29,11 \times 23,13 \times 19\} ; \\
A_{4}=\{2 \times 37,3 \times 31,5 \times 29,7 \times 23,11 \times 19,13 \times 17\} ; \\
A_{5}=\{2 \times 31,3 \times 29,5 \times 23,7 \times 19,11 \times 17\} ; \\
A_{6}=\{2 \times 29,3 \times 23,5 \times 19,7 \times 17,11 \times 13\} .
\end{array}
$$

It is easy to see that $A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6}$ are pairwise disjoint, and any 5 elements in $A_{i}(i=1,2,3,4,5,6)$ are pairwise coprime, $\left|A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5} \cup A_{6}\right|=88$.

If we arbitrarily select 217 elements from $S$, then at most $280-88=192$ elements do not belong to $A_{1} \cup A_{2} \cup A_{3} \cup A_{4} \cup A_{5} \cup A_{6}$, so at least $217-192=25$ elements are taken from $\bigcup_{i=1}^{6} A_{i}$. By the pigeonhole principle, at least $\left[\frac{25-1}{6}\right]+1=5$ elements come from the same drawer, and these 5 elements meet the requirement. Therefore, the smallest $n \leqslant 217$.
In summary, the smallest $n=217$.