For $n \leqslant 100$, we have $f(n)=91$.
(1983 Yugoslav Olympiad)
1. Let $n \leqslant 100$ and $n+11>100$, i.e., $90 \leqslant n \leqslant 100$, thus, $f(n)=f[f(n+11)]=f(n+11-$ 10) $=f(n+1)$. Therefore, $f(90)=f(91)=\cdots=f(100)=f(101)=91$.

Let $n<90$, take $m \in \mathbf{N}$, such that $90 \leqslant n+11 m \leqslant 100$, then we have $f(n)=f^{(2)}(n+11)=\cdots=f^{(m+1)}(n+11 m)=$ $f^{(m)}(f(n+11 m))=f^{(m)}(91)=91$. This proves that for any $n \leqslant 100$, we have $f(n)=91$.