Example 3 For each positive integer $n$, find the largest real number $C(n)$ satisfying the following inequality,
$$
C(n) \leqslant\left|\frac{f(a)-f(b)}{a-b}\right|,
$$

where $f$ is a polynomial of degree $n$ and $f: \mathbf{Z} \rightarrow \mathbf{Z}$, $a, b \in \mathbf{Z}, f(a) \neq f(b)$
【Analysis】Given that $f$ is an integer-valued polynomial of degree $n$, from Example 2 we know that
$[1,2, \cdots, n] \frac{f(a)-f(b)}{a-b} \in \mathbf{Z}$.
Thus, $\left|\frac{f(a)-f(b)}{a-b}\right| \geqslant \frac{1}{[1,2, \cdots, n]}$.
Next, we prove that there exists such an integer-valued polynomial, making $C(n)=\frac{1}{[1,2, \cdots, n]}$ achievable.
For convenience, let $M^{-1}=\frac{1}{[1,2, \cdots, n]}$.
Since $f$ is an integer-valued polynomial, there exist $a_{i} \in \mathbf{Z}$ $(i=1,2, \cdots, n)$, such that $f(x)=\sum_{k=0}^{n} a_{k}\binom{x}{k}\left(a_{0}=0\right)$.
Therefore, $f(0)=0$.
Let $b=0$, then the problem is equivalent to finding an integer $a$ such that
$$
\left|\frac{f(a)-f(0)}{a-0}\right|=\frac{1}{M}.
$$

For simplicity, let $a$ be $M$, then the problem is equivalent to finding $a_{i} \in \mathbf{Z}(i=1,2, \cdots, n)$, such that $f(M)=1$, i.e.,
$$
\sum_{k=1}^{n} a_{k}\binom{M}{k}=1.
$$

By Bézout's theorem, it suffices to show that
$$
\operatorname{gcd}\left(\binom{M}{1},\binom{M}{2}, \cdots,\binom{M}{n}\right)=1,
$$

where $M=[1,2, \cdots, n]$.
Suppose $\operatorname{gcd}\left(\binom{M}{1},\binom{M}{2}, \cdots,\binom{M}{n}\right)=d>1$. If there is a prime $p$ such that $p \mid d$, then $p \left\lvert\,\binom{M}{1}\right.$.
Since $d \left\lvert\,\binom{M}{1}\right.$, it follows that $p \mid M$. Hence, $p \leqslant n$.
Let $v_{p}(n)=a$. Then $d\left|\binom{M}{p^{a}} \Rightarrow p\right|\binom{M}{p^{a}}$.
But $M=M_{0} p^{a}$, which contradicts Lucas' theorem.