Four, (16 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$. Find the number of elements in the set
$$
\left\{n \left\lvert\, n=\left[\frac{k^{2}}{2005}\right]\right., 1 \leqslant k \leqslant 2004, k \in \mathbf{N}\right\}
$$
From $\frac{(k+1)^{2}}{2005}-\frac{k^{2}}{2005}=\frac{2 k+1}{2005} \leqslant 1$, we solve to get
$k \leqslant 1002$.
When $k=1,2, \cdots, 1002$, we have
$$
\left[\frac{(k+1)^{2}}{2005}\right]=\left[\frac{k^{2}}{2005}\right]
$$

or $\left[\frac{(k+1)^{2}}{2005}\right]=\left[\frac{k^{2}}{2005}\right]+1$.
Since $\left[\frac{1002^{2}}{2005}\right]=500,\left[\frac{1^{2}}{2005}\right]=0$, thus, when $k=1,2, \cdots, 1002$, $\left[\frac{k^{2}}{2005}\right]$ can take all values of $0,1, \cdots, 500$.

Additionally, when $k=1003,1004, \cdots, 2004$, since $\frac{(k+1)^{2}}{2005}-\frac{k^{2}}{2005}>1$, we have
$$
\left[\frac{(k+1)^{2}}{2005}\right] \geqslant\left[\frac{k^{2}}{2005}\right]+1,
$$

which means $\left[\frac{1003^{2}}{2005}\right],\left[\frac{1004^{2}}{2005}\right], \cdots,\left[\frac{2004^{2}}{2005}\right]$ are all distinct. There are $2004-1002=1002$ such numbers.
Noting that $\left[\frac{1003^{2}}{2005}\right]=501>\left[\frac{1002^{2}}{2005}\right]$, we know that the set
$$
\left\{n \left\lvert\, n=\left[\frac{k^{2}}{2005}\right]\right., 1 \leqslant k \leqslant 2004, k \in \mathbf{N}\right\}
$$

has a total of $501+1002=1503$ elements.