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 $11 \le x_1 \le x_2$).
later It should be easy to find solutions with $k=n$ although I have no idea about the integer case: Pick $y_1,\cdots,y_n$ not too small and no two too close together (say $y_i=i$) Then the values $y^j_1 + y^j_2 + \cdots + y^j_n$ determine the coefficients of the monic polynomial $f(t)=\prod_1^n(t-y_i)$ and vice versa. The $y_i$ are the $n$ roots of $f$. Now pick $x_{n+1},\cdots,x_{2n}$ positive but "small enough".Then the desired equations $x^j_1 + x^j_2 + \cdots + x^j_n=y^j_1 + y^j_2 + \cdots + y^j_n-\sum_1^nx_{n+k}^j$ determine the coefficents of some monic polynomial $g(t)$ whose roots are $x_1,\cdots,x_n$. If the prechosen values are small enough (maybe $x_{n+k}=\frac{k}{100^n}$) then the coeffcients of $g$ should be only slightly preturbed from those of $f$ so the roots $x_1,\cdots,x_n$ should be only slightly preturbed from $y_1,\cdots,y_n$
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.