The polynomial

$c_0 + c_1x + \cdots + c_{n-2}x^{n-2} + x^n(x-y_1)(x-y_2)\cdots(x-y_n)$

with all $y_i > 0$ can only have $2n$ positive real zeros if its $n-2$th derivative has $n+2$ positive real zeros. The $n-2$th derivative is

$(n-2)!c_{n-2} + \frac{(2n)!}{(n+2)!}x^2(x-y_1')\cdots(x-y_n')$,

where the $y_i'$s are all positive. This will have $n+2$ real zeros iff $c_{n-2} \ge 0$, but if $c_{n-2} \gt 0$ then one of the roots will be negative. If $c_{n-2} = 0$ and $c_j \ne 0$ for some $j \lt n-2$, then by looking at the $j$th derivative (and assuming $j$ is maximal such that $c_j \ne 0$) we see that we don't even get $2n$ real roots of the original polynomial.

Thus, you can't find any solutions in the positive reals with $k \ge n+1$. Aaron gave a sketch of a reason that you can expect to find solutions with $k = n$.

The polynomial

$c_0 + c_1x + \cdots + c_{n-2}x^{n-2} + x^n(x-y_1)(x-y_2)\cdots(x-y_n)$

with all $y_i > 0$ can only have $2n$ positive real zeros if its $n-2$th derivative has $n+2$ positive real zeros. The $n-2$th derivative is

$(n-2)!c_{n-2} + \frac{(2n)!}{(n+2)!}x^2(x-y_1')\cdots(x-y_n')$,

where the $y_i'$s are all positive. This will have $n+2$ real zeros iff $(-1)^n c_{n-2} \le 0$, but if $(-1)^n c_{n-2} \lt 0$ then one of the roots will be negative. If $c_{n-2} = 0$ and $c_j \ne 0$ for some $j \lt n-2$, then by looking at the $j$th derivative (and assuming $j$ is maximal such that $c_j \ne 0$) we see that we don't even get $2n$ real roots of the original polynomial.

Thus, you can't find any solutions in the positive reals with $k \ge n+1$. Aaron gave a sketch of a reason that you can expect to find solutions with $k = n$.