Seven, (25 points) Let $n$ be a positive integer, $a=[\sqrt{n}]$ (where $[x]$ denotes the greatest integer not exceeding $x$). Find the maximum value of $n$ that satisfies the following conditions:
(1) $n$ is not a perfect square;
(2) $a^{3} \mid n^{2}$.
(Zhang Tongjun
Zhu Yachun, problem contributor)
Seven, from (1) we get $a<\sqrt{n}<a+1$, then $a^{2}<n<a^{2}+2 a+1$,
that is
$a^{2}+1 \leqslant n \leqslant a^{2}+2 a$.
Let $n=a^{2}+t(t \in\{1,2, \cdots, 2 a\})$.
From (2) we have
$a^{3}\left|\left(a^{4}+2 a^{2} t+t^{2}\right) \Rightarrow a^{2}\right| t^{2} \Rightarrow a \mid t$.
Furthermore, $a^{3} \mid \left(a^{4}+2 a^{2} t+t^{2}\right) \Rightarrow a^{3} \mid t^{2}$.
Let $t^{2}=k a^{3}$, then $t=a \sqrt{k a}$.
Since $t, a, k \in \mathbf{N}_{+}$, we have $\sqrt{k a} \in \mathbf{N}_{+}$.
From $t \in\{1,2 ; \cdots, 2 a\}$, we have $t=a \sqrt{k a} \leqslant 2 a$, that is $\sqrt{k a} \leqslant 2$.
Therefore, $\sqrt{k a}=1$ or $2, k a \leqslant 4, a \leqslant 4$.
Since $n=a^{2}+t$, and $a \leqslant 4, t \leqslant 2 a$, we can let $a=4, t=2 a=8$. Then $n=a^{2}+t=16+8=24$ is the maximum.
Upon verification, $n=24$ satisfies conditions (1) and (2).
Therefore, the maximum value of $n$ is 24.