For which $n$ can we divide $1,2,\ldots,2n$ into $n$ pairs so that the product of the sum of the $n$ pairs is a perfect square?
If $n$ is even, this is possible: match the first number with the last, second number with the second-to-last, etc., so every pair has sum $2n+1$ and the product is $(2n+1)^n$. This is a perfect square because $n$ is even.
For $n=1$ it is clearly impossible, but for $n=3$ it is possible: $(1+5)(2+4)(3+6) = 18^2$.
Best Answer
You're almost done just take for $n$ odd $(1+5)(2+4)(3+6)(7+2n)(8+2n-1)\cdots=6^2\cdot 3^2\cdot (2n+7)^{n-3}$