3 (Full score: 50 points) For an integer $n \geqslant 4$, find the smallest integer $f(n)$, such that for any positive integer $m$, in any $f(n)$-element subset of the set $\{m, m+1, \cdots, m+n-1\}$, there are at least 3 pairwise coprime elements.
Solution 1 When $n \geqslant 4$, consider the set $M=\{m, m+1, m+2, \cdots, m+n-1\}$.
If $2 \mid m$, then $m+1, m+2, m+3$ are pairwise coprime;
If $2 \nmid m$, then $m, m+1, m+2$ are pairwise coprime.
Thus, in all $n$-element subsets of $M$, there are at least 3 elements that are pairwise coprime.

Therefore, $f(n)$ exists, and $f(n) \leqslant n$.
Let $T_{n}=\{t \mid t \leqslant n+1$ and $2 \mid t$ or $3 \mid t\}$, then $T_{n}$ is a subset of $\{2,3, \cdots, n+1\}$, but any 3 elements in $T_{n}$ cannot be pairwise coprime, so $f(n) \geqslant \left|T_{n}\right|+1$.
By the principle of inclusion and exclusion, we have
$$
\left|T_{n}\right|=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right],
$$

Thus, it must be that
$$
f(n) \geqslant\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1 . \quad \text { (1) } \cdots 20 \text { points }
$$

Therefore,
$$
\begin{array}{l}
f(4) \geqslant 4, f(5) \geqslant 5, f(6) \geqslant 5, \\
f(7) \geqslant 6, f(8) \geqslant 7, f(9) \geqslant 8 .
\end{array}
$$

The following proves that $f(6)=5$.
Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be 5 numbers in $\{m, m+1, \cdots, m+5\}$. If these 5 numbers include 3 odd numbers, then they are pairwise coprime; If these 5 numbers include 2 odd numbers, then there must be 3 even numbers. Without loss of generality, assume $x_{1}, x_{2}, x_{3}$ are even, and $x_{4}, x_{5}$ are odd. When $1 \leqslant i4$, in the full set of five elements, three numbers can be chosen that are pairwise coprime, so $f(5)=5$.
For $n=6: \{m, m+1, m+2, m+3, m+4, m+5\}$ contains three odd and three even numbers. If a four-element subset of three even and one odd is taken, then there are no three numbers that are pairwise coprime, so $f(6)>4$. Considering a five-element subset, if there are three odd numbers, then these three numbers must be pairwise coprime; if there are three even and two odd numbers, since at most one of the three even numbers is divisible by 3, and at most one is divisible by 5, there must be one even number that is not divisible by 3 or 5. This number is pairwise coprime with the other two odd numbers, so $f(6)=5$.
For $n>6$: Let $T_{n}=\{t: t \leqslant n+1, 2 \mid t$ or $3 \mid t\}$, then $T_{n}$ is a subset of $\{2,3, \cdots, n+1\}$, and any three elements in $T_{n}$ cannot be pairwise coprime, so
$$
\begin{array}{c}
F(n) \geqslant\left|T_{n}\right|+1=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1 . \\
\text { Let } n=6 k+r, k \geqslant 1, r=0,1,2,3,4,5, \text { then } \\
{\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1} \\
=4 k+\left[\frac{r+1}{2}\right]+\left[\frac{r+1}{3}\right]-\left[\frac{r+1}{6}\right]+1 .
\end{array}
$$

It is easy to verify that $\left[\frac{r+1}{2}\right]+\left[\frac{r+1}{3}\right]-\left[\frac{r+1}{6}\right]=\left\{\begin{array}{l}r, r=0,1,2,3, \\ r-1, r=4,5 .\end{array}\right.$
When $r=0,1,2,3$, divide the $n=6 k+r$ numbers into $k$ groups as $\{m, m+1, \cdots, m+5\},\{m+6, m+7, \cdots, m+11\}, \cdots,\{m+6(k-1), m+6 k-$
$5, \cdots, m+6 k-1\}$, and leave $r$ numbers $m+6 k, \cdots, m+6 k+$ $r-1.4 k+r+1$ numbers, at least $4 k+1$ numbers are divided into $k$ drawers, at least one drawer has five numbers, since $f(6)=5$, there must be three numbers that are pairwise coprime. When $r=4,5$, it can be similarly proved that there must be three numbers that are pairwise coprime. Therefore, $f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1$.