7. In the sequence
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]
$$

there are $\qquad$ distinct integers ( $[x]$ denotes the greatest integer not exceeding the real number $x$).
7.1515.

Let the $k$-th term of the known sequence be $\left[\frac{k^{2}}{2019}\right]$. Then, when $(k+1)^{2}-k^{2} \leqslant 2019$, i.e., $k \leqslant 1009$,
$$
\begin{array}{l}
\frac{(k+1)^{2}}{2019}=\frac{k^{2}}{2019}+\frac{2 k+1}{2019} \leqslant \frac{k^{2}}{2019}+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right] \leqslant 1+\left[\frac{k^{2}}{2019}\right] \\
\Rightarrow 0 \leqslant\left[\frac{(k+1)^{2}}{2019}\right]-\left[\frac{k^{2}}{2019}\right] \leqslant 1 .
\end{array}
$$

Since $\left[\frac{1009^{2}}{2019}\right]=504$, therefore, in
$$
\left[\frac{1^{2}}{2019}\right],\left[\frac{2^{2}}{2019}\right], \cdots,\left[\frac{1009^{2}}{2019}\right]
$$

the integers $0,1, \cdots, 504$ appear without omission, and these 505 integers are all distinct.
$$
\begin{array}{l}
\text { Also, when } k \geqslant 1010, \\
2 k+1=(k+1)^{2}-k^{2}>2019 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right]=\left[\frac{k^{2}+2 k+1}{2019}\right] \\
\quad \geqslant\left[\frac{k^{2}+2019}{2019}\right]=\left[\frac{k^{2}}{2019}\right]+1 \\
\Rightarrow\left[\frac{(k+1)^{2}}{2019}\right]-\left[\frac{k^{2}}{2019}\right] \geqslant 1 .
\end{array}
$$

Thus, $\left[\frac{1010^{2}}{2019}\right],\left[\frac{1011^{2}}{2019}\right], \cdots,\left[\frac{2019^{2}}{2019}\right]$ are
distinct integers, and all are greater than 504.
Therefore, the number of distinct integers in the original sequence is
$$
505+1010=1515 .
$$