First note \[\begin{aligned}1+\frac{1}2+\frac{1}6+...+\frac{1}{n(n+1)} & = 1-\frac{1}2+\frac{1}2-\frac{1}3+...+\frac{1}n-\frac{1}{n+1} \\ & =1-\frac{1}{n+1} \\ & =\frac{n}{n+1}\end{aligned}\] Now since $N$ is divided into $n$ parts inversely proportional to 2,6,12..., we have for some constant $x$ that \[N=x+\frac{x}2+\frac{x}6+...+\frac{x}{n(n+1)}\] The problem states that the smallest part is $\frac{N}{400}$. That means $\frac{x}{n(n+1)}=\frac{N}{400}$ or $x=\frac{Nn(n+1)}{400}$. Now we can say that \[\begin{aligned}N & =x+\frac{x}2+\frac{x}6+...+\frac{x}{n(n+1)} \\ & =\left(1+\frac{1}2+\frac{1}6+...+\frac{1}{n(n+1)}\right)x \\ & =\left(\frac{n}{n+1}\right)\left(\frac{Nn(n+1)}{400}\right) \\ & =\frac{Nn^2}{400} \\ 1 & =\frac{n^2}{400} \\ n^2 & =400 \\ n & =\boxed{20}\end{aligned}\]