[Solution] Factorization yields
$$
\begin{array}{l}
4^{27}+4^{500}+4^{n} \\
=4^{27}\left(4^{n-27}+4^{473}+1\right), \\
4^{n-27}+4^{473}+1=\left(2^{n-27}\right)^{2}+2 \cdot 2^{945}+1 .
\end{array}
$$

When $n-27=945$, i.e., $n=972$, (2) is a perfect square, and thus (1) is a perfect square.
When $n>972$,
$$
\begin{aligned}
& \left(2^{n-27}+1\right)^{2} \\
= & \left(2^{n-27}\right)^{2}+2 \cdot 2^{n-27}+1 \\
> & \left(2^{n-27}\right)^{2}+2 \cdot 2^{945}+1 \\
> & \left(2^{n-27}\right)^{2} .
\end{aligned}
$$

Therefore, at this time, (2) is not a perfect square, and thus (1) is not a perfect square.
Hence, the maximum value of $n$ is 972.
2 - 87 Let $4^{27}+4^{500}+4^{n}$ be a square number (the square of an integer), find the maximum value of the integer $n$.

