Solution by Daniel Vaz, member of the Portuguese IMO team in 2007 and 2008: Suppose there isn't $ k$ with $ 1\leq k \leq 2002$ for which $ a_k<0$. First, observe that $ a_n$ is decreasing for $ 1\leq n \leq 2002$, because $ a_{n+1}=a_n-\frac{1}{a_n} \leq a_n$, since $ a_n\geq 0$. Now, note that for $ n\geq1$, $ a_n\leq a_1 =56$, so $ a_{n+1}\leq a_n -\frac{1}{56}$, and from this we easily get that $ a_{57}\leq 56-\frac{56}{56}=55$. Now in a similar manner we obtain $ a_n\leq 55$ for $ n\geq 1+56$, and then get $ a_{57+55}\leq 54$, and continuing this we get $ a_{1597}=a_{1+56+55+...+1} \leq 56-56=0$, contradiction, as desired. In fact we proved that $ a_k\leq 0$ for some $ k$, with $ 1\leq k\leq 1957$, which is something a bit stronger

Jorge Miranda wrote: Now, note that for $ n\geq1$, $ a_n\leq a_1 = 56$, so $ a_{n + 1}\leq a_n - \frac {1}{56}$, and from this we easily get that $ a_{57}\leq 56 - \frac {56}{56} = 55$. I thought that if we substitute $ 56$ for $ n$ we get $ a_{57} \leq a_{56} - 1/56 \leq 56 - 1/56.$

No, you don't plug it in like that, but instead use it repeatedly (or induction ), like this: $ a_{57}\leq a_{56} -\frac{1}{56} \leq a_{55}-\frac{2}{56}\leq \dots \leq a_1-\frac{56}{56}=55$ All the other times are analogous, of course

Oh, I didn't think that way; thanks.