For the integer $n \geqslant 4$, find the smallest integer $f(n)$, such that for any positive integer $m$, any $f(n)$-element subset of the set $\{m, m+1, m+2, \cdots, m+n-1\}$ contains at least 3 pairwise coprime elements.
$$
\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1
$$

Three, [Analysis and Solution] When $n \geqslant 4$, for 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 does not divide $m$, then $m, m+1, m+2$ are pairwise coprime.
$\therefore$ In all $n$-element subsets of $M$, there are at least 3 elements that are pairwise coprime, i.e., $f(n)$ exists, and $f(n) \leqslant 0$.
Let $T_{n}=\{t \mid t \leqslant n+1$ and $2 \mid t$, or $3 \mid t\}$, it is known that $T_{n}$ is a subset of $\{2,3, \cdots, n+1\}$, but any 3 elements in $T_{n}$ cannot be pairwise coprime, $\therefore f(n) \geqslant\left|T_{n}\right|+1$.
By the principle of inclusion and exclusion, $\left|T_{n}\right|=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]$,
thus, it must be that $f(n) \geqslant\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1$
$$
\therefore f(4) \geqslant 4, f(5) \geqslant 5, f(6) \geqslant 5, f(7) \geqslant 6, f(8) \geqslant 7, f(9) \geqslant 8 \text {. }
$$

We will prove that $f(6)=5$.
Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be 5 numbers in $\{m, m+1, m+2, \cdots, m+5\}$, 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, let $x_{1}, x_{2}, x_{3}$ be even, and $x_{4}, x_{5}$ be odd. When $1 \leqslant i<j \leqslant 3$, $\left|x_{i}-x_{j}\right| \in\{2,4\}$, so at most one of $x_{1}, x_{2}, x_{3}$ is divisible by 3, and at most one is divisible by 5, thus at least one is neither divisible by 3 nor by 5. Without loss of generality, let it be $x_{3}$, then $x_{3}, x_{4}, x_{5}$ are pairwise coprime. This shows that there are 3 numbers that are pairwise coprime, i.e., $f(6)=5$.
Also, since $\{m, m+1, m+2, \cdots, m+n\}=\{m, m+1, m+2, \cdots, m+n-1\} \cup\{m+n\}$,
it is known that $f(n+1) \leqslant f(n)+1$.
$$
\because f(6)=5, \therefore f(4)=4, f(5)=5, f(7)=6, f(8)=7, f(9)=8 \text {. }
$$
$\therefore$ When $4 \leqslant n \leqslant 9$, $f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1$
Next, we use induction to prove that (2) holds for all $n$:
Assume that (2) holds for $n \leqslant k(k \leqslant 9)$,
When $n=k+1$, since
$$
\{m, m+1, m+2, \cdots, m+k\}=\{m, m+1, m+2, \cdots, m+k-6\} \cup\{m+k-5, m+k-4, m+k-3, m+k-2, m+k-1, \cdots, m+k\},
$$
and when $n=6, n=k-5$, (2) holds,
$$
\therefore f(k+1) \leqslant f(k-5)+f(6)-1=\left[\frac{k+2}{2}\right]+\left[\frac{k+2}{3}\right]-\left[\frac{k+2}{6}\right]+1
$$

By (1), (3), for $n=k+1$, (2) holds.
$\therefore$ For any $n \geqslant 4, f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1$.