$6 \cdot 2$ A function $f$ defined on the positive integers satisfies:
$$
f(n)=\left\{\begin{array}{l}
n-3 \\
f(f(n+7))
\end{array}\right.
$$
(when $n \geqslant 1000$),
(when $n<1000$).
Find $f(90)$.
[Solution] Let $f^{m}(n)=\underbrace{f(f \cdots f}_{m \text { times }}(n) \cdots)$.
Since $90+7 \times 130=1000$,

we have $f(90)=f^{2}(97)=\cdots=f^{131}(1000)$.
Also, $f(1000)=997$;
$$
\begin{array}{l}
f(999)=f(f(1006))=f(1003)=1000 ; \\
f(998)=f(f(1005))=f(1002)=999 ; \\
f(997)=f(f(1004))=f(1001)=998 .
\end{array}
$$

Therefore, the value of $f^{m}(1000)$ cycles through $997 \rightarrow 998 \rightarrow 999 \rightarrow 1000 \rightarrow \cdots$, with a period of 4.
Since $131 \div 4=32$ remainder 3,
we have $f(90)=f^{131}(1000)=f^{3}(1000)=999$.