Let $S=\{1,2,3, \ldots, 280\}$. Determine the smallest integer $n$ such that every $n$-element subset of $S$ contains 5 numbers that are pairwise relatively prime.
Let's select from the set $S$ the multiples of 2, 3, 5, and 7. Among these, no matter how we choose 5 numbers, there will be two that are multiples of the same prime, and thus not relatively prime. The number of elements selected from $S$ is calculated using the sieve formula to be 216. Therefore, the integer $n$ we are looking for is at least 217.

We will prove that for $n=217$, the condition is already satisfied. For this, we need to examine how many elements can be in a subset of $S$ where there are no 5 numbers that are pairwise relatively prime. In the selection of the 216-element set mentioned above, we excluded all primes and 1 except for four, all prime squares except for four, and additionally six numbers that can be written as the product of two primes:

$$
\begin{array}{llllll}
11 \cdot 13 & 11 \cdot 23 & 11 \cdot 17 & 13 \cdot 17 & 11 \cdot 19 & 13 \cdot 19
\end{array}
$$

In any set that does not meet the original condition, there can be at most 4 primes and 1, otherwise there would be 5 numbers that are pairwise relatively prime, so at least as many must be excluded as in the previous example. The same applies to prime squares. All that remains is to ensure that at least 6 more elements must be excluded. Below, I provide 6 sets of 5 elements each, which are subsets of $S$, do not contain primes or prime squares, and in each set, the numbers are pairwise relatively prime, so at least one element of each set must be excluded from $S$ for the condition not to be met.

The sets are:

$$
\begin{array}{ll}
\left\{2^{3} ; 3^{3} ; 5 \cdot 23 ; 7 \cdot 19 ; 11 \cdot 13\right\} & \left\{2^{6} ; 3 \cdot 17 ; 5^{3} ; 7 \cdot 13 ; 11 \cdot 23\right\} \\
\left\{2^{4} ; 3^{4} ; 5 \cdot 19 ; 7 \cdot 23 ; 11 \cdot 17\right\} & \left\{2^{7} ; 3 \cdot 23 ; 5 \cdot 29 ; 7 \cdot 11 ; 13 \cdot 19\right\} \\
\left\{2^{5} ; 3^{5} ; 5 \cdot 13 ; 7 \cdot 17 ; 11 \cdot 19\right\} & \left\{2^{8} ; 3 \cdot 19 ; 5 \cdot 11 ; 7 \cdot 29 ; 13 \cdot 17\right\}
\end{array}
$$

Therefore, if we select a 217-element subset from $S$, it will either contain one of the 6 sets mentioned earlier, or it will contain 5 primes, or it will contain 5 prime squares. In any of these cases, there will be 5 numbers that are pairwise relatively prime. Thus, the number $n$ we are looking for is 217.