20.3.5 ** 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.
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-exclusion, $\left|T_{n}\right|=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]$, thus we must have
$$
f(n) \geqslant\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1 \text {, }
$$

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 $1 \leqslant i<j \leqslant 3$, $\left|x_{i}-x_{j}\right| \in\{2,4\}$, 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, assume $3 \nmid x_{3}, 5 \nmid x_{3}$, then $x_{3}, x_{4}, x_{5}$ are pairwise coprime. This means that among these 5 numbers, there are 3 that are pairwise coprime, i.e., $f(6)=5$.

Furthermore, by $\{m, m+1, \cdots, m+n\}=\{m, m+1, \cdots, m+n-1\} \cup\{m+n\}$, we know $f(n+1) \leqslant f(n)+1$.
Since $f(6)=5$, we have
$$
f(4)=4, f(5)=5, f(7)=6, f(8)=7, f(9)=8 .
$$

Therefore, when $4 \leqslant n \leqslant 9$,
$$
f(n)=\left[\frac{n+1}{2}\right]+\left[\frac{n+1}{3}\right]-\left[\frac{n+1}{6}\right]+1 .
$$

The following uses induction to prove (2) holds for all $n$:
Assume (2) holds for $n \leqslant k(k \geqslant 9)$.
When $n=k+1$, since
$$
\begin{aligned}
\{m, m+1, \cdots, m+k\}= & \{m, m+1, \cdots, m+k-6\} \cup \\
& \{m+k-5, m+k-4, m+k-3, \\
& m+k-2, m+k-1, m+k\},
\end{aligned}
$$

and by the induction hypothesis for $n=6, n=k-5$, (2) holds, so
$$
\begin{aligned}
f(k+1) & \stackrel{*}{\leqslant} f(k-5)+f(6)-1 \\
& =\left[\frac{k+2}{2}\right]+\left[\frac{k+2}{3}\right]-\left[\frac{k+2}{6}\right]+1 .
\end{aligned}
$$

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