Problem 1631

1. Define Good and Bad Pairs:
We define a pair $(i, j)$ as "good" if $i \cdot j$ is a perfect square, and "bad" otherwise. Let our initial numbers be $a_1, a_2, \ldots, a_{2n}$. Suppose we have a non-square matching $S_1 = (a_1, a_2), S_2 = (a_3, a_4), \ldots, S_n = (a_{2n-1}, a_{2n})$.

2. Key Claim:
If we take any two pairs, say the first two pairs, then either both $(a_1, a_3)$ and $(a_2, a_4)$ are good, or both $(a_1, a_4)$ and $(a_2, a_3)$ are good.

3. Proof of the Claim:
Assume the contrary. Without loss of generality (WLOG), suppose $(a_1, a_3)$ is bad. If both $(a_1, a_3)$ and $(a_1, a_4)$ are bad, then $(a_3, a_4)$ must be good, which contradicts the assumption that $(a_3, a_4)$ is part of a non-square matching. Similarly, if both $(a_1, a_3)$ and $(a_2, a_3)$ are bad, then $(a_1, a_2)$ must be good, which again contradicts the assumption. Therefore, the claim is proved.

4. Induction Hypothesis:
We will use induction to prove the statement. The base case for $n = 1$ is trivial since there is only one pair, and it is given to be a non-square matching.

5. Inductive Step:
Assume the statement is true for $n-1$. From the claim, for each good pair $S_i, S_j$, we can create new good pairs $S'_i, S'_j$ such that $S_i \cup S_j = S'_i \cup S'_j$.

6. Constructing New Matchings:
Let's make a new match for $a_1$ from the set $S_i$, $i \neq 1$. The remaining numbers can form at least $(n-1)!$ matchings. Thus, we get $(n-1) \cdot (n-1)!$ new matchings, which equals $n! - (n-1)!$.

7. Fixing Pairs and Continuing:
Now fix $S_1$ and do the same for $a_3$. Similarly, we will get at least $(n-1)! - (n-2)!$ new matchings. Continue this process until the end. So we will get at least $(n! - (n-1)!) + ((n-1)! - (n-2)!) + \cdots + (2! - 1!) = n! - 1$ new matchings.

8. Adding the Initial Matching:
Including the initial matching $S_1, S_2, \ldots, S_n$, we get at least $n! - 1 + 1 = n!$ matchings.

$\blacksquare$

The final answer is $\boxed{ n! }$