One can't have $k \ge 2n$ (proof in a moment).

If $k \le n$ then one can choose $x_1,x_2,\cdots,x_n$ and $y_{n+1},y_{n+2},\cdots,y_{2n}$ and solve for $y_1,y_2,\cdots,y_n$. I arbitrarily decided to try this with $y_3=3,y_4=4$ Varying $x_1,x_2$ I find

$x_1,x_2;y_1,y_2,y_3,y_4=8,20;\frac{21-\sqrt{437}}{2},\frac{21+\sqrt{437}}{2},3,4$ Many other choices work as well (for example $11 \le x_1 \le x_2$).

Here is my argument for why we can't expect $k=2n$: In this case the equations and values for $x_1,...,x_n$ will determine $y_1,y_2,\cdots ,y_{2n}$ up to order. But we know a solution with $n$ zeros so the other solutions must be the same rearranged.