(i) Let $n=p_{1}^{a(1)} \cdots p_{r}^{a(r)}$, we have
$$\begin{aligned} k & =d\left(n^{2}\right) / d(n) \\ & =(1+2 a(1)) \cdots(1+2 a(r)) /[(1+a(1)) \cdots(1+a(r))] \end{aligned}$$

Thus, $k$ must be an odd number. Therefore, $n$ must be a perfect square.
(ii) Now we prove that any odd number $k$ can be expressed in this way. We use induction. When $k=1$, we can take $n=1$; when $k=3$, we can take $n=\left(2 \cdot 3^{2}\right)^{2}$. Hence, the conclusion holds for $k=1,3$. Assume that for all positive odd numbers $k \leqslant 4 N+3(N \geqslant 0)$, the conclusion holds. We will prove that for all positive odd numbers $k \leqslant 4(N+1)+3$, the conclusion holds. This requires considering two cases.
(a) When $k=4(N+1)+1$, by the induction hypothesis and the fact that $n$ must be a perfect square, there exists $n_{1}$ such that $1+2(N+1)=d\left(n_{1}^{4}\right) / d\left(n_{1}^{2}\right)$. Now take $n=n_{1}^{2} p^{2(N+1)}$, where the prime $p$ is coprime with $n_{1}$. We then have
$$k=4(N+1)+1=d\left(n^{2}\right) / d(n)$$
(b) When $k=4(N+1)+3$, we have $k=2^{l} m-1,(2, m)=1, l \geqslant 2$. By the induction hypothesis and the fact that $n$ must be a perfect square, there exists $n_{1}$ such that $m=d\left(n_{1}^{4}\right) / d\left(n_{1}^{2}\right)$. Take distinct primes $p_{1}, \cdots, p_{l}$, all coprime with $n_{1}$, and let $2 \beta(j)=3^{j} 2^{l-j} m-2(1 \leqslant j \leqslant l-1)$, $2 \beta(l)=3^{l-1} m-1$, $n=n_{1} p_{1}^{\beta(1)} \cdots p_{l}^{\beta(l)}$. We then have
$$k=4(N+1)+3=d\left(n^{2}\right) / d(n)$$

Therefore, for all positive odd numbers $k \leqslant 4(N+1)+3$, the conclusion holds. Proof completed.