$$
\begin{array}{l}
{\left[\frac{1^{2}}{1980}\right],\left[\frac{2^{2}}{1980}\right],\left[\frac{3^{2}}{1980}\right], \cdots,} \\
{\left[\frac{1900^{2}}{1980}\right] \text { How many different numbers are there in the sequence? }}
\end{array}
$$
Solution: First, note that when $\alpha-\beta>1$, the value of $[a] 1_{j} [\beta]$ is definitely different. Therefore, we solve the inequality
$$
\frac{(k+1)^{2}}{1980}-\frac{k^{2}}{1980}>1,
$$

which simplifies to
$$
\begin{array}{l}
2 k+1>1980, \\
k>989 .
\end{array}
$$

Thus, starting from the 990th term, these 1980 terms, i.e.,
$$
\left[\frac{990^{2}}{1980}\right],\left[\frac{991^{2}}{1980}\right], \cdots,\left[\frac{1980^{2}}{1980}\right]
$$

are all different, totaling 991 distinct numbers.
On the other hand, when $k \leqslant 44$,
$$
\begin{array}{l}
0<\frac{k^{2}}{1980}<1, \\
{\left[\frac{k^{2}}{1980}\right]=0 .}
\end{array}
$$

Thus, the first 44 numbers are all 0.
When $45 \leqslant k \leqslant 62$,
$$
\begin{array}{l}
1<\frac{k^{2}}{1980}<2, \\
{\left[\frac{k^{2}}{1980}\right]=1 .}
\end{array}
$$

Therefore, the first 62 numbers can only be 0 or 1.
Also, since
$$
\left[\frac{989^{2}}{1980}\right]=494, \quad\left[\frac{990^{2}}{1980}\right]=495,
$$

the first 989 terms must include the 495 distinct numbers $0,1,2, \cdots, 494$.
Thus, there are $991+495=1486$ distinct numbers.