Iberoamerican Mathematical Olympiad

Solution:

Note that whilst $a_{n}$ remains positive we have $a_{1}>a_{2}>a_{3}>\ldots>a_{n}$. Hence if $a_{m}$ and $a_{m+n}$ are in this part of the sequence, then $a_{m+1}=a_{m}-1/a_{m}$, $a_{m+2}=a_{m+1}-1/a_{m+1}<a_{m+1}-1/a_{m}=a_{m}-2/a_{m}$. By a trivial induction $a_{m+n}<a_{m}-n/a_{m}$.

If we use one step then we need $56^{2}=3136$ terms to get $a_{1+3136}<56-56^{2}/56=0$, which is not good enough. So we try several steps.

Thus suppose that $a_{n}>0$ for all $n\leq 2002$. Then we get successively:

$a_{337}<56-336/56=50$

$a_{837}<50-500/50=40$

$a_{1237}<40-400/40=30$

$a_{1537}<30-300/30=20$

$a_{1737}<20-200/20=10$

$a_{1837}<10-100/10=0$.

Contradiction. So we must have $a_{n}<0$ for some $n<2002$.