Problem 1195

[Solution] For natural number $n$, we discuss by cases.
Case $1$: $n$ is odd.
In this case, $f(n)=2$, so $l_{n}=1$.
Case $2$: $n$ is even.
In this case, we set $n=2^{a}(2 m+1)$, where $a \geqslant 1, m \geqslant 0$.
If all odd numbers $t$ satisfying $1<t<2^{a+1}$ are divisors of $n$, then
$$\begin{array}{l} f(n)=2^{a+1}, \\ f(f(n))=3, \\ f(f(f(n)))=2 . \end{array}$$

Therefore, $\quad l_{n}=3$.
If not all odd numbers $t$ satisfying $1<t<2^{a+1}$ are divisors of $n$, then there must be a smallest positive odd number $t_{0}\left(1<t_{0}<2^{a+1}\right)$ that is not a divisor of $n$. Thus we have
$$\begin{array}{l} f(n)=t_{0}, \\ f(f(n))=f\left(t_{0}\right)=2 . \end{array}$$

Therefore, $l_{n}=2$.
In summary, we have
$$l_{n}=\left\{\begin{array}{l} 1, n \text { is odd, } \\ 3, n \text { is even } 2^{a}(2 m+1), a \geqslant 1, m \geqslant 0, \\ \quad \text { and all odd numbers } t \text { satisfying } 1<t<2^{a+1} \text { are divisors of } n, \\ 2, n \text { is even } 2^{a}(2 m+1), a \geqslant 1, m \geqslant 0, \\ \text { and there exists an odd number } t_{0} \text { that is not a divisor of } n, \text { satisfying } 1<t_{0}<2^{a+1} . \end{array}\right.$$