Answer: D

Small Hint:

If $N_2$ is a multiple of $N_1$ then $f_1(N_2)$ is a multiple of $f_1(N_1),$ so unboundedness is inherited by multiples; find the minimal “essential” $N$

Big Hint:

An essential $N$ has every exponent at least $2$ and at most two prime factors; analyze powers of $2,3$ first, then the few remaining prime powers and prime pairs below $400$

Solution:

If $N_2$ is a multiple of $N_1$ then $f_1(N_2)$ is a multiple of $f_1(N_1),$ so if $S_{N_1}$ is unbounded so is $S_{N_2}.$ Call $N$ essential if it is unbounded but no proper divisor is. An essential $N$ must have all exponents at least $2,$ and $(p_1\cdots p_k)^2\le400$ forces at most two primes.

For $n=2^a3^b,$ two iterations send the exponent pair to $(2a-4,2b-3).$ Thus the orbit is unbounded exactly when $a\ge5$ or $b\ge4,$ producing the essential values $2^5=32$ and $3^4=81.$ For a single other prime, the bound leaves only $5^2,5^3,7^2,7^3,11^2,13^2,17^2,19^2;$ direct application of $f_1$ leaves only $7^3=343$ essential. With two primes, $p_1p_2\le20$ leaves the pairs $(2,5),(2,7),(3,5)$ besides $(2,3).$ Applying $f_1$ to their possible squareful products leaves only $2^4\cdot5^2=400.$

Their multiples up to $400$ number $\lfloor\frac{400}{32}\rfloor=12,$ $\lfloor\frac{400}{81}\rfloor=4,$ $\lfloor\frac{400}{343}\rfloor=1,$ and $\lfloor\frac{400}{400}\rfloor=1,$ with no overlaps, for a total of $12+4+1+1=18.$

Thus, the correct answer is D.