Let $(p, q)=1, q=2^{\alpha} \cdot 5^{\beta} \cdot d$, where $\alpha \geqslant 0, \beta \geqslant 0, (d, 10)=1$. Then the length of the repeating decimal part of $\frac{p}{q}$, denoted as $s$, should be the smallest positive integer satisfying $10^{s} \equiv 1(\bmod d)$. Given $s=n$. Since $10^{n} \equiv 1(\bmod d)$, we have $10^{n}=1+k d$, where $k$ is a positive integer. Therefore, $10^{n\left(10^{n}-1\right)}=(1+k d)^{10^{n}-1}=1+k_{1} d^{2}$, where $k_{1}$ is an integer, i.e., $10^{n\left(10^{n}-1\right)} \equiv 1\left(\bmod d^{2}\right)$. Since $q^{2}=2^{2 \alpha} 5^{2 \beta} d^{2}$, the length of the repeating decimal part of $\left(\frac{p}{q}\right)^{2}$, denoted as $s_{0}$, satisfies $s_{0} \leqslant n\left(10^{n}-1\right)$.

On the other hand, this upper bound can be achieved. Let $d=9$, clearly $n=1$, and the smallest $s$ satisfying $10^{s} \equiv 1\left(\bmod 9^{2}\right)$ is $9=n\left(10^{n}-1\right)$. Therefore, the maximum length of the repeating decimal part of $\left(\frac{p}{q}\right)^{2}$ is $n\left(10^{n}-1\right)$.

Comment: When $d=10^{n}-1$, the repeating decimal part of $\frac{1}{d}$ has a length of $n$. And the repeating decimal part of $\frac{1}{d^{2}}$ has a length of $n\left(10^{n}-1\right)$. In fact, if there is $10^{m} \equiv 1\left(\bmod d^{2}\right)$, then $10^{m} \equiv 1(\bmod d)$. Since $n$ is the smallest natural number satisfying $10^{n} \equiv 1(\bmod d)$, we have $m=k n$. $10^{m}-1=10^{k n}-1=(d+1)^{k}-1 \equiv k d\left(\bmod d^{2}\right)$, thus $d\left|k, n d=n\left(10^{n}-1\right)\right| m$.
19.2.17 ** The decimal representation of the fraction $\frac{p}{q}(p, q \in \mathbf{N})$ has a repeating cycle of length $n$. What is the maximum possible length of the repeating cycle in the decimal representation of the fraction $\left(\frac{p}{q}\right)^{2}$?