$14 \cdot 36$ In the sequence $\left[\frac{1^{2}}{1980}\right],\left[\frac{2^{2}}{1980}\right],\left[\frac{3^{2}}{1980}\right], \cdots,\left[\frac{1980^{2}}{1980}\right]$, how many different numbers are there?
(6th All-Russian Mathematical Olympiad, 1980)
[Solution] Let $x_{k}=\left[\frac{k^{2}}{1980}\right]$.
Then the sequence $\left\{x_{k}\right\}$ is a non-decreasing sequence.
(1) When $k \leqslant 44$, by $44^{2}=1936 < 1980$,
$$
2 \cdot 1980
$$
So when $45 \leqslant k \leqslant 62$,
$$
11$ when, $x_{k+1}$ and $x_{k}$ are obviously different.
At this point, by $2 k+1>1980$ solving for $k>989$.
Thus, when $989<k \leqslant 1980$, all $x_{k}$ are different, that is, $x_{990}, x_{991}, \cdots, x_{1980}$ these 991 numbers are different.
(4) When $k=989$,
$$
\left[\frac{989^{2}}{1980}\right]=494 \text {. }
$$

When $k \leqslant 989$, $y_{k}<1$, at this time $x_{k}$ and $x_{k+1}$ are either the same or differ by 1, so in $x_{1}, x_{2}, \cdots, x_{989}$, there must be the different integers $0,1,2, \cdots, 494$.
Therefore, in the known sequence $\left\{x_{k}\right\}$, the total number of different numbers that appear is
$$
991+495=1486 \text { (numbers). }
$$