There must exists prime divisor $p$ of $n$ that $\nu_p (n)$ is odd. Let $\nu_p (n)=2k+1$ for some non-negative $k$. We then pair up $(dp^i,dp^{k+1+i})$ for each divisor $d$ of $n/p^{2k+1}$ and non-negative integer $i\leqslant k$. Clearly, this result in the partition of all divisors of $n$ into pairs as desired.

Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} \dots p_{s}^{\alpha_{s}}$ denote the canonical representation of a positive integer $n$. We give a proof by induction on $s$. For $s=1$, we have $n=p_{1}^{\alpha_{1}}$. As $n$ is not a perfect square, $\alpha_{1}$ must be odd so we can take pairs $(1,p_{1}), (p_{2}, p_{3}), \dots, (p_{1}^{\alpha_{1}-1}, p_{1}^{\alpha_{1}})$. Assume now that the claim holds for all non-perfect squares having up to $k$ distinct prime factors (i.e., $s \leq k$). We prove that it also holds for all non-perfect squares having $s=k+1$ distinct prime factors. Let $n=p_{1}^{\alpha_{1}}p_{2}^{\alpha_{2}} \dots p_{k}^{\alpha_{k}} p_{k+1}^{\alpha_{k+1}}$ be any such number. Since $n$ is not a perfect square, there exists at least one $i \in \{1,2, \dots, k+1\}$ such that $\alpha_{i}$ is odd. Let $n=m\cdot p_{j}^{\alpha_{j}}$, where $j \neq i$. As $p_{i}^{\alpha_{i}}$ is a factor in canonical representation of $m$, we have that $m$ is not a perfect square. Since $m$ has exactly $k$ distinct prime factors, we can apply the inductive hypothesis to obtain partition of all divisors of $m$ into pairs $(d_{1}, d_{2}), (d_{3}, d_{4}), \dots, (d_{\tau(m)-1}, d_{\tau(m)})$ where the first number in each pair divides the second. To obtain such partition of divisors of $n$, first observe that divisors of $n$ can be partitioned into two sets: (i) divisors not divisible by $p_{j}$ and (ii) divisors divisible by $p_{j}$. Divisors from (i) represent exactly the set of divisors of $m$ and thus they can be partitioned into desired pairs in the same way as assumed above. For divisors from (ii), observe that they are equal to the following union of sets $\cup_{t=1}^{\alpha_{j}}\{p_{1}^{t}d_{1}, p_{1}^{t}d_{2}, \dots, p_{1}^{t}d_{\tau(m)}\}$. They can be combined in pairs $(p_{j}^{t}d_{1}, p_{j}^{t}d_{2}), (p_{j}^{t}d_{3}, p_{j}^{t}d_{4}), \dots, (p_{j}^{t}d_{\tau(m)-1}, p_{j}^{t}d_{\tau(m)})$ for $t=1,2, \dots, \alpha_{j}$ and this pairing satisfies the condition that one of the divisors in the pair divides the other. As we have clearly covered all divisors of $n$, our proof is completed. COMMENT 1: By taking $t$ to go from $0$ to $\alpha_{j}$ we can slightly shorten the above proof as this would eliminate the need for discussing cases (i) and (ii) COMMENT 2 Note that in the pairing for $m$ we do not assume that $d_{1} < d_{2} < \dots < d_{\tau(m)}$, as is typically assumed in most problems that involve divisors of a positive integer.