For $k = 1$, $S(k) = 1$. For $k > 1$, note that for a positive integer $a < k$ relatively prime to $k$ the number $k - a$ is also relatively prime to $k$. Hence the summands in the sum $S(k)$ can be paired and each pair has sum $k$. The number of positive integers relatively prime to $k$ (and less than $k$) is $\varphi(k)$, so
$$S(k) = \frac{k\varphi(k)}{2}.$$
Let $q$ be the largest prime factor of $m$, and let $2 = p_1 < p_2 < \dots < p_s = q$ be consecutive prime numbers. Then $m = p_1^{a_1} \dots p_s^{a_s}$ (where some of $a_i$ are 0). We will construct a number $x$ of the form $x = p_1^{b_1} \dots p_s^{b_s}$ satisfying the condition of the problem. Note that
$$\begin{aligned} 2S(x) &= x\varphi(x) = p_1^{b_1} \dots p_s^{b_s} \cdot p_1^{b_1-1} (p_1-1) \dots p_s^{b_s-1} (p_s-1) \\ &= p_1^{2b_1-1} (p_1-1) \dots p_s^{2b_s-1} (p_s-1). \end{aligned}$$
Also, for every $i \le s$ all prime divisors of $p_i - 1$ are among $p_1, \dots, p_s$, so we choose $c_1, \dots, c_s$ such that $(p_1 - 1) \dots (p_s - 1) = p_1^{c_1} \dots p_s^{c_s}$, i.e.
$$2S(x) = p_1^{2b_1+c_1-1} \dots p_s^{2b_s+c_s-1}.$$
The number $x$ satisfies the condition of the problem if and only if $2b_i + c_i - 1$ is divisible by $n$ and $b_i \ge a_i$, for all $i \in \{1, \dots, s\}$.
Since $n$ is odd, its multiples alternate in being even and odd, so for every $i$ we can choose a large enough $k_i \in \mathbb{N}$ such that $k_i n \equiv c_i - 1 \pmod 2$ and
$$b_i = \frac{k_i n - c_i + 1}{2} \ge a_i.$$

Then $2S(x) = p_1^{2b_1+c_1-1} \dots p_s^{2b_s+c_s-1} = (p_1^{k_1} \dots p_s^{k_s})^n$, which shows that $x$ satisfies the condition of the problem.