A cool name

Notice that $200 = 2^3 * 5^2$ . Thus we can say that the $N$ will be in the form $a_1^{x_1} * a_2^{x_2} * a_3^{x_3} * \dots$ such that $a_k$ is prime and $(x_1 + 1) * (x_2 + 1) * (x_3 + 1) * \dots=200$ (so that $N$ has $200$ factors by the product law). We need to minimise $N$ , and this occurs when $N = 2^4 * 3^4 * 5 * 7 * 11 = 498960$ since $(4 + 1) * (4+1) * (1+1)* (1+1)* (1+1) = 200$ . Hence, the digit sum of $N$ is $4 + 9 + 8 + 9 + 6 + 0 = \fbox{36}$