For each positive integer $n$, let $c(n)$ be the largest real number such that
$$c(n) \le \left| \frac{f(a) - f(b)}{a - b} \right|$$
for all triples $(f, a, b)$ such that:
- $f$ is a polynomial of degree $n$ taking integers to integers, and
- $a, b$ are integers with $f(a) \neq f(b)$.

To find $c(n)$, we claim that $c(n) = \frac{1}{L_n}$, where $L_n = \text{lcm}(1, 2, 3, \ldots, n)$.

First, note that any polynomial $f(X)$ that maps the integers to the integers can be represented as:
$$f(X) = c_0 + c_1 \binom{X}{1} + c_2 \binom{X}{2} + \cdots + c_n \binom{X}{n}.$$

### Lemma 1
$L_n \cdot \frac{\binom{a}{n} - \binom{b}{n}}{a - b} \in \mathbb{Z}$.

Proof:
Consider the polynomial $g(X) = \binom{X + b}{n} - \binom{b}{n}$. This polynomial can be written as:
$$g(X) = d_1 \binom{X}{1} + \cdots + d_n \binom{X}{n}.$$
Using the identity $\frac{1}{X} \binom{X}{n} = \frac{1}{n} \binom{X-1}{n-1}$, the denominator of $\frac{g(X)}{X}$ must have size at most $L_n$. Thus, $L_n \cdot \frac{g(X)}{X} \in \mathbb{Z}$, proving the lemma. $\blacksquare$

Now, consider:
$$T = \frac{f(a) - f(b)}{a - b} = \sum_{k=0}^n c_k \frac{\binom{a}{k} - \binom{b}{k}}{a - b}.$$
In particular, for each prime $p$,
$$v_p \left( c_k \frac{\binom{a}{k} - \binom{b}{k}}{a - b} \right) \ge -v_p(L_k) \ge -v_p(L_n),$$
so $v_p(T) \ge -v_p(L_n)$. Therefore, $T \cdot L_n \in \mathbb{Z}$. If $T \neq 0$, then $T \ge \frac{1}{L_n}$, establishing a lower bound on $c(n)$.

To show that this lower bound is attainable, consider a suitable choice of $c_i$ such that:
$$\frac{f(N!) - f(0)}{N!} = \frac{1}{L_n}$$
for large $N$. Note that:
$$\frac{\binom{N!}{k} - \binom{0}{k}}{N! - 0} = \frac{\binom{N!}{k}}{N!} = \frac{\binom{N! - 1}{k - 1}}{k}.$$
No prime less than or equal to $k$ divides $\binom{N! - 1}{k - 1}$, as the expression can be written as $\prod_{i=1}^{k-1} \frac{N! - i}{i}$ and $\gcd \left( \frac{N! - i}{i}, L_k \right) = 1$ for large $N$ and $k \le n$. Therefore:
$$\frac{f(N!) - f(0)}{N! - 0} = \sum_{k=0}^n \frac{c_k t_k}{k}$$
for $\gcd(t_k, k) = 1$ fixed and some $c_k$. By Bézout's identity, we can choose suitable $c_i$ such that the expression equals $\frac{1}{L_n}$.

Thus, we conclude that:
$$c(n) = \frac{1}{L_n}.$$

The answer is: $\boxed{\frac{1}{L_n}}.$