*7. Find all natural numbers $n \in [1000, 2000]$ 
such that
$$
a_{0}=\sqrt{57121+35 n}
$$

is a natural number.
Solving based on the condition $1000 \leqslant n \leqslant 2000$, we get
$$
a_{0} \geqslant \sqrt{57121+35 \cdot 1000}=303.51441 \cdots
$$

and $a_{n} \leqslant \sqrt{57121}+35 \cdot 2000=356.54032 \cdots$, i.e., $304 \leqslant a_{a} \leqslant 356$.

Since $57121=1632 \cdot 35+1$, we have $a_{\mathrm{n}}=\sqrt{35 i n+1632}+1$, i.e., $a_{\mathrm{n}}^{2}-1=\left(a_{0}-1\right)\left(a_{n}+1\right)$ can be divided by $5=5 \cdot 7$. Therefore, either $7 \mid\left(a_{0}-1\right)$ or $7 \mid\left(a_{0}+1\right)$. If $a_{4}=7k+1$, then $a_{n}=7k-1$. Thus, $44 \leqslant k \leqslant 50$. Using a calculator for $k=44, 45, \cdots, 50$ to compute $a_{\mathrm{n}}^{2}-1=(7 k+1)^{2}-1$, and selecting those divisible by 5, we get four values of $n=\left[\left(a_{n}^{2}-1\right)/35\right]-1632$: $1096\left(k=44, a_{0}=209\right), 1221$ $\left(k=45, a_{\mathrm{n}}=316\right), 1749\left(k=49, a_{\mathrm{n}}=344\right)$, and $1888\left(k=50, a_{\mathrm{a}}=351\right)$.

In the second case, $a_{a}=7 k-1$ and $45 \leqslant k \leqslant 51$. Similarly, we get four values of $n$: $1185(k=45, a_{n}=314)$, $1312\left(k=46, a_{\mathrm{a}}=321\right)$, 
$$
\begin{array}{l}
1848\left(k=50, a_{n}=319\right) \quad 1999\left(k=51, a_{0}=356\right).
\end{array}
$$