Let $d(n)$ denote the number of positive integers that divide $n$, including $1$ and $n$. For example, $d(1)=1,d(2)=2,$ and $d(12)=6$. (This function is known as the divisor function.) Let\[f(n)=\frac{d(n)}{\sqrt [3]n}.\]There is a unique positive integer $N$ such that $f(N)>f(n)$ for all positive integers $n\ne N$. What is the sum of the digits of $N?$
$\textbf{(A) }5 \qquad \textbf{(B) }6 \qquad \textbf{(C) }7 \qquad \textbf{(D) }8\qquad \textbf{(E) }9$
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 $1\leq i\leq k,$ the denominator grows faster than the numerator, as exponential functions always grow faster than polynomial functions. Therefore, for each prime $p_i$ with $\left(p_1,p_2,p_3,p_4,\ldots\right)=\left(2,3,5,7,\ldots\right),$ 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 $N=2^3\cdot3^2\cdot5^1\cdot7^1=2520.$ The sum of its digits is $2+5+2+0=\boxed{\textbf{(E) }9}.$
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 $\textbf{(E)}$ is possible.
~MRENTHUSIASM