I claim that for all even $n$ there exist numbers, and for odd $n$ there don't.

Indeed, let's explore the case $n = 2$. Due to Cauchy-Schwarz inequality

$(11x_{1}^{2}+12y_{1}^{2})(11x_{2}^{2}+12y_{2}^{2})\geq (11x_{1}x_{2} + 12y_{1}y_{2})^{2}$ with equality if and only if $\frac{x_{1}}{x_{2}}=\frac{y_{1}}{y_{2}}= k$ with constant $k$. We take the equality case as it's an integer

and a perfect square. Then we have infinite quadruples $(x_{1},x_{2},y_{1},y_{2})$ for which it's a perfect square.

For all even $n$, we can divide it in many products of two numbers, each of which is a perfect square.

Thus, for even $n$ we can accomplish our goal.

It only remains to prove that it's impossible for odd $n$. Suppose there exists tuples $(x_{1},.....,x_{n},y_{1},.....,y_{n})$, and take the one with the least sum. Let's look at the product modulo $3$. We have, $2x_{1}^{2}\cdot .....\cdot 2x_{n}^{2}\equiv 2^{n}\cdot x_{1}^{2}\cdot ...\cdot x_{n}^{2}$

$\equiv 2\cdot x_{1}^{2}\cdot ...\cdot x_{n}^{2}\equiv c^{2}(mod 3)$

It's trivial that $x^{2}\equiv 0,1(mod 3)$. Then, if $3\not |x_{i}$ for all $i$, we have that ,

$2\cdot x_{1}^{2}\cdot ...\cdot x_{n}^{2}\equiv 2 (mod 3)$, which is not a quadratic residue. Then $3|x_{i}$ for some $i$, and so $x_{i} = 3a_{i}$.

Now, let's analyze the factor $(11x_{i}^{2}+12y_{i}^{2})$ . It's clear that $3\not |y_{i}$, because if it divided it we could have the same tuple with $\frac{x_{i}}{3}$ and $\frac{y_{i}}{3}$,, which would still be a perfect square but would have less sum. Then $3\parallel (11x_{i}^{2} +12y_{i}^{2})$.

As the product must be a perfect square, we must have that $3$ divides the product of all other factors except $11x_{i}^{2}+12y_{i}^{2}$ . Looking modulo $3$ we have $3|x_{k}$ for some $k$, and so $x_{k} = 3a_{k}$. We have again that $3\parallel (11x_{k}^{2}+12y_{k}^{2})$. Consider the whole product, but divide the factors $(11x_{i}^{2}+12y_{i}^{2})$ and $(11x_{k}^{2}+12y_{k}^{2})$ by $3$. We have:

$(11x_{1}^{2}+12y_{1}^{2})...(11\cdot 3a_{i}^{2} + 4y_{i}^{2})...(11\cdot 3a_{k}^{2}+4y_{k}^{2})...(11x_{n}^{2}+12y_{n}^{2}=c'^{2}$.

Looking modulo $3$ again, we will see $3$ divides some $x_{\ell}$, with $\ell \neq i,k$. If we continue with this procedure,we will always find pairs of numbers $(x_{a},x_{b})$ such that $3$ divides both of them and we discard them.But as $n$ is odd, there will always remain a number in the end. We've reached a contradiction and we're

done.