Let $n = 2^a n'$, where $a$ is a non-negative integer and $n'$ is odd, and let $2^b$ be the highest power of $2$ dividing $\sigma(n')$. Let $p_1, \dots, p_\ell$ be odd primes not dividing $n'$ (e.g., let each $p_i > n'$), and let $N$ be an integer such that $r = N\varphi(p_1^2 \cdots p_\ell^2 n') - 1 > \max(a, b)$, where $\varphi$ is Euler's totient function. If $t$ is one of the $2^\ell$ divisors of the product $p_1 \cdots p_\ell$, and $s = 2^r n' t^2$, then $s$ is a multiple of $n$, since $r > a$, and
$$\sigma(s) = \sigma(2^r)\sigma(n')\sigma(t^2) = (2^{r+1}-1) \cdot 2^b \cdot \text{odd},$$
since $\sigma(t^2)$ is odd. Hence,
$$2^{r-b} \frac{\sigma(s)}{s} = \frac{2^{r+1}-1}{n' t^2} \cdot \text{odd}$$
is an odd integer, by Euler's theorem. Finally, since there are $2^\ell$ such $s$, one for each divisor $t$ of the product $p_1 \cdots p_\ell$, it is sufficient to consider an integer $\ell \ge \log_2 m$ to produce a set $S$ of at least $m$ multiples of $n$ satisfying the required condition; plainly, $k = r - b$ does not depend on $s$.