For each $m \in \mathbb{N}$, let $f(m)$ be the product of the primes that appear with odd exponent in the prime factorization of $m$. It is easy to see that given two positive integers $a$ and $b$, the product $ab$ is a perfect square if and only if $f(a) = f(b)$.

For each $k \in \mathbb{N}$, let $S$ be the set of all $m$ in $\{1, 2, \dots, 4n\}$ such that $f(m) = k$. Note that if $k$ is not squarefree, then $S$ is empty. Also, as $f(m) \leq m$ for all $m$, we have that each element of $\{1, 2, \dots, 4n\}$ belongs to exactly one of the sets $S_1, S_2, \dots, S_{4n}$.

By the initial observation each of the pairs that we are going to form must be integrated by two elements of the same $S_j$; furthermore, any way of assembling the pairs respecting that rule meets the conditions of the statement. Then, if each set $S_j$ has $a_j$ elements, the maximum number of pairs that can be formed is $\lfloor \frac{a_1}{2} \rfloor + \lfloor \frac{a_2}{2} \rfloor + \dots + \lfloor \frac{a_{4n}}{2} \rfloor$.

We claim that $\lfloor \frac{a_j}{2} \rfloor$ is equal to the number of multiples of $4$ in $S_j$. Indeed, this is obvious if $S_j$ is empty; if instead $j$ is squarefree, the elements of $S_j$ are the numbers of the form $j \cdot s^2$ with $s$ covering the values from $1$ to $a_j$. Since $j$ is squarefree, it has at most one factor of $2$, so $j \cdot s^2$ is a multiple of $4$ if and only if $s$ is even. Then, the number of multiples of $4$ in $S_j$ coincides with the number of even numbers between $1$ and $a_j$, which is precisely $\lfloor \frac{a_j}{2} \rfloor$.

From the above, the sum $\lfloor \frac{a_1}{2} \rfloor + \lfloor \frac{a_2}{2} \rfloor + \dots + \lfloor \frac{a_{4n}}{2} \rfloor$ matches the number of multiples of $4$ among all the sets $S_1, S_2, \dots, S_{4n}$, that is, the number of multiples of $4$ in $\{1, 2, \dots, 4n\}$. This quantity is $n$, and that is the answer to the problem.