IMO Hk TST

The greatest $n$ is $145$.

One can easily check that
$$145 \mapsto 42 \mapsto 20 \mapsto 4 \mapsto 16 \mapsto 37 \mapsto 58 \mapsto 89 \mapsto 145,$$
so that $f^8(145) = 145$. It remains to check $n \le 145$.

We first prove that $n$ has at most $3$ digits. Suppose on the contrary that $n = \overline{a_1a_2\cdots a_k}$ where $k \ge 4$. For any $m \le n$, we have
$$f(m) \le f(\underbrace{99\cdots9}_{k \text{ times}}) = 81k < 10^{k-1} \le n.$$
(To prove $81k < 10^{k-1}$ for $k \ge 4$, we check the base case $324 < 1000$ and the inductive step $81(k+1) < 10(81k) < 10^k$.) Using this, we can easily prove that $f^s(n) < n$ for any $s \in \mathbb{Z}^+$. Thus, it is impossible that $f^s(n) = n$.

Similarly, if $n > 243$, then for $m \le n$, we have
$$f(m) \le f(999) = 243 < n.$$
If $163 < n \le 243$, then for $m \le n$, we have
$$f(m) \le f(199) = 163 < n.$$
Thus, we must have $n \le 163$.

Note that $n$ is also the image of some term in the cycle. Let $n = f(t)$ for some $t \le 163$. If $n > 145 = 8^2 + 9^2$, then the only possibility is $t = 99$. But then we have
$$99 \mapsto 162 \mapsto 41 \mapsto 17 \mapsto 50 \mapsto 25 \mapsto 29 \mapsto 85 \mapsto 89 \mapsto 145,$$
which enters the cycle with largest term $145$, and so it never goes back to $162$. Therefore, we must have $n \le 145$. This completes the proof.