1. Let's consider the members of the finite sequence

$$
\left[\frac{k^{2}}{1998}\right], \quad k=1,2, \ldots, 1997
$$

where $[x]$ denotes the integer part of $x$. How many different members are there in this sequence?

(Greek)
1. Consecutive members of the sequence $\frac{1^{2}}{1998}, \frac{2^{2}}{1998}, \ldots, \frac{999^{2}}{1998}$ differ by less than 1, so among their integer parts, all integers from $\left[\frac{1^{2}}{2014}\right]=0$ to $\left[\frac{999^{2}}{1998}\right]=499$ appear. On the other hand, among the 999 numbers $\frac{999^{2}}{1998}, \frac{1000^{2}}{1998}, \ldots, \frac{1997^{2}}{1998}$, each two differ by more than 1, so their integer parts are distinct. Thus, in the given sequence, we have a total of $500+999-1=1498$ distinct members.