jmc

Denote by (1) and (2) the two parts of the definition of $f$, respectively. If we begin to use the definition of $f$ to compute $f(84)$, we use (2) until the argument is at least $1000$: $$f(84)=f(f(89))=f(f(f(94)))=\cdots =f^N(1004)$$(where $f^N$ denotes composing $f$ with itself $N$ times, for some $N$). The numbers $84,89,94,\dots ,1004$ form an arithmetic sequence with common difference $5$; since $1004-84=920=184\cdot 5$, this sequence has $184+1=185$ terms, so $N=185$.

At this point, (1) and (2) are both used: we compute $$\begin{array}{rl}{f}^N(1004)& \stackrel{(1)}{=}{f}^{N-1}(1001)\stackrel{(1)}{=}{f}^{N-2}(998)\stackrel{(2)}{=}{f}^{N-1}(1003)\stackrel{(1)}{=}{f}^{N-2}(1000)\\ & \stackrel{(1)}{=}{f}^{N-3}(997)\stackrel{(2)}{=}{f}^{N-2}(1002)\stackrel{(1)}{=}{f}^{N-3}(999)\stackrel{(2)}{=}{f}^{N-2}(1004).\end{array}$$Repeating this process, we see that $$f^N(1004)=f^{N-2}(1004)=f^{N-4}(1004)=\cdots =f^3(1004).$$(The pattern breaks down for $f^k(1004)$ when $k$ is small, so it is not true that $f^3(1004)=f(1004)$.) Now, we have $$f^3(1004)\stackrel{(1)}{=}{f}^2(1001)\stackrel{(1)}{=}f(998)\stackrel{(2)}{=}{f}^2(1003)\stackrel{(1)}{=}f(1000)\stackrel{(1)}{=}\boxed{997}.$$