$7 \cdot 73$ Let $n \in N$, and make $37.5^{n}+26.5^{n}$ a positive integer, find the value of $n$.
(Shanghai Mathematical Competition, 1998)
[Solution] $37.5^{n}+26.5^{n}=\frac{1}{2^{n}}\left(75^{n}+53^{n}\right)$.
When $n$ is a positive even number,
$$
75^{n}+53^{n} \equiv(-1)^{n}+1^{n} \equiv 2(\bmod 4),
$$

at this time $75^{n}+53^{n}$ is a number of the form $4l+2$, thus $37.5^{n}+26.5^{n}$ cannot be a positive integer.
When $n$ is a positive odd number,
$$
\begin{aligned}
75^{n}+53^{n} & =(75+53)\left(75^{n-1}-75^{n-2} \times 53+\cdots+53^{n-1}\right) \\
& =2^{7}\left(75^{n-1}-75^{n-2} \times 53+\cdots+53^{n-1}\right) .
\end{aligned}
$$

The expression inside the parentheses has $n$ terms, and each term is odd, so the sum inside the parentheses is the sum of an odd number of odd numbers, which is odd.
Therefore, the positive odd number $n$ can only be $n=1,3,5,7$.
From the above, we know that $37.5^{n}+26.5^{n}$ is a positive integer only when $n=1,3,5,7$.