smc

We consider the prime factorization of $n:$ $$n=\prod _{i=1}^k{p}_i^{e_i}.$$ By the Multiplication Principle, we have $$d(n)=\prod _{i=1}^k(e_i+1).$$ Now, we rewrite $f(n)$ as $$f(n)=\frac{d(n)}{\sqrt[3]{n}}=\frac{{\prod }_{i=1}^k(e_i+1)}{{\prod }_{i=1}^k{p}_i^{e_i/3}}=\prod _{i=1}^k\frac{e_i+1}{p_i^{e_i/3}}.$$ As $f(n)>0$ for all positive integers $n,$ note that $f(a)>f(b)$ if and only if $f(a{)}^3>f(b{)}^3$ for all positive integers $a$ and $b.$ So, $f(n)$ is maximized if and only if $$f(n{)}^3=\prod _{i=1}^k\frac{(e_i+1{)}^3}{p_i^{e_i}}$$ is maximized. For each independent factor $\frac{(e_i+1{)}^3}{p_i^{e_i}}$ with a fixed prime $p_i,$ where $1\le i\le k,$ the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. Therefore, for each prime $p_i$ with $\left(p_1,p_2,p_3,p_4,\dots \right)=\left(2,3,5,7,\dots \right),$ we look for the nonnegative integer $e_i$ such that $\frac{(e_i+1{)}^3}{p_i^{e_i}}$ is a relative maximum: \begin{array}{c|c|c|c|c} & & & & \\ [-2.25ex] \boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{\dfrac{(e_i+1)^3}{p_i^{e_i}}} & \textbf{Max?} \\ [2.5ex] \hline\hline & & & & \\ [-2ex] 1 & 2 & 0 & 1 & \\ & & 1 & 4 & \\ & & 2 & 27/4 &\\ & & 3 & 8 & \checkmark\\ & & 4 & 125/16 & \\ [0.5ex] \hline & & & & \\ [-2ex] 2 & 3 & 0 & 1 &\\ & & 1 & 8/3 & \\ & & 2 & 3 & \checkmark\\ & & 3 & 64/27 & \\ [0.5ex] \hline & & & & \\ [-2ex] 3 & 5 & 0 & 1 & \\ & & 1 & 8/5 & \checkmark\\ & & 2 & 27/25 & \\ [0.5ex] \hline & & & & \\ [-2ex] 4 & 7 & 0 & 1 & \\ & & 1 & 8/7 & \checkmark\\ & & 2 & 27/49 & \\ [0.5ex] \hline & & & & \\ [-2ex] \geq5 & \geq11 & 0 & 1 & \checkmark \\ & & \geq1 & \leq8/11 & \\ [0.5ex] \end{array} Finally, the positive integer we seek is $N=2^3\cdot 3^2\cdot 5^1\cdot 7^1=2520.$ The sum of its digits is $2+5+2+0=\boxed{9}.$ Alternatively, once we notice that $3^2$ is a factor of $N,$ we can conclude that the sum of the digits of $N$ must be a multiple of $9.$ Only choice $\text{(E)}$ is possible.