One can't have $k \ge 2n$ (proof in a moment).
If $k \le n$ then one can choose $y_1,y_2,\cdots,y_n$ and $x_{n+1},x_{n+2},\cdots,x_{2n}$ and solve for $x_1,x_2,\cdots,x_n$. I arbitrarily decided to try this with $x_3=3,x_4=4$ Varying $y_1,y_2$ I find
$y_1,y_2;x_1,x_2,x_3,x_4=8,20;\frac{21-\sqrt{437}}{2},\frac{21+\sqrt{437}}{2},3,4$ Many other choices work as well (for example $11 \le x_1 \le x_2$).
I'd guess that if one starts with $y_1,\cdots,y_n$ (all larger than 1 say) and also sets $x_{n+1},\cdots,x_{2n}$ positive but very small then there will be $x_1,\cdots,x_n$ which are a small perturbation of the $y$ values.
Here is my argument for why we can't expect $k=2n$: In this case the equations and values for $y_1,...,y_n$ will determine $x_1,x_2,\cdots ,x_{2n}$ up to order. But we know a solution with $n$ zeros so the other solutions must be the same rearranged.