Since
$$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)} = \frac{n}{n+1},$$
we guess that $f(n) = n$ is the unique solution satisfying the given condition.
Substituting $n = 1$, we get $f(f(1))f(1) = 1$, so $f(1) = 1$.
Substituting $n+1$ for $n$ in the given condition, we get
$$\frac{f(f(n))}{f(n+1)} + \frac{1}{f(n+1)f(n+2)} = \frac{f(f(n+1))}{f(n+2)}.$$
The above is equivalent to
$$f(f(n))f(n+2) + 1 = f(f(n+1))f(n+1).$$
Note: $f(n+1) = 1 \Rightarrow f(f(n+1)) = 1$. Therefore, $f(f(n))f(n+2) = 0$ cannot occur. Hence $f(n) > 1, \forall n > 1$.
Using mathematical induction, we prove: $f(f(n)) < f(n+1)$.
When $n = 1$, $f(2) > 1 = f(f(1))$ holds.
If $f(n+1) > f(f(n))$ then $f(n+1) \ge f(f(n)) + 1$. Therefore
$$f(f(n))f(n+2) + 1 \ge f(f(n+1))f(f(n)) + f(f(n+1)).$$
Since $n+1 > 1$, we have $f(n+1) > 1$. That is, $f(f(n+1)) > 1$. From this we can deduce
$$f(n+2) > f(f(n+1)).$$

$f(n + 1) > f(f(n)), \forall n \in N.$
$$\text{Below we prove that the function satisfying (1) is } f(n) = n, \forall n \in N.$$
Let the set $S = \{f(f(1)), f(2), f(f(2)), f(2), \dots, f(f(n-1)), f(n), f(f(n)), f(n+1), \dots\}$, then the set $S$ has a smallest element $f(n_0)$, for some positive integer $n_0$. By (1), we can obtain $f(n_0) = 1$. Below we prove: $n_0 = 1$.
Suppose $n_0 > 1$ then $1 = f(n_0) > f(f(n_0 - 1))$, which is a contradiction! Therefore $f(1) = 1$ and $f(n) > 1$ for $n > 1$.
Consider $f: \{n \ge 2\} \to \{n \ge 2\}$, similarly we can obtain: $f(2) = 2$ and $f(n) > 2$ for $n > 2$. By mathematical induction we can prove:
$$f(k) = k, \text{ and } f(n) > k, \text{ for } n > k,$$
therefore the unique solution satisfying the problem is $f(n) = n, \forall n \in N$.