First of all, we have $\lfloor\sqrt{n}\rfloor=\lceil\sqrt{n}\rceil$ when $n$ is a perfect square and $\lfloor\sqrt{n}\rfloor=\lceil\sqrt{n}\rceil-1$ otherwise. Therefore, in the first case, the original equation holds if and only if $\sqrt{n}$ is a perfect square itself, i.e., $n$ is a fourth power. In the second case, we need $m=\lfloor\sqrt{n}\rfloor$ to satisfy the equation $\lfloor\sqrt{m+1}\rfloor=\lceil\sqrt{m}\rceil$, which happens if and only if either $m$ or $m+1$ is a perfect square $k^{2}$. Therefore, $n$ is magical if and only if $\left(k^{2}-1\right)^{2}<n<\left(k^{2}+1\right)^{2}$ for some (positive) integer $k$. There are $\left(k^{2}+1\right)^{2}-\left(k^{2}-1\right)^{2}=4 k^{2}-1$ integers in this range. The range in the problem statement includes $k=1,2, \ldots, 9$ and the interval $\left(99^{2}, 100^{2}\right]$, so the total number of magical numbers is $$4\left(1^{2}+2^{2}+\cdots+9^{2}\right)-9+\left(100^{2}-99^{2}\right)=4 \cdot \frac{9 \cdot(9+1) \cdot(18+1)}{6}+190=1330$$