A positive integer $N$ is divided in $n$ parts inversely proportional to the numbers $2, 6, 12, 20, \ldots$
The smallest part is equal to $\frac{1}{400} N$. Find the value of $n$.
1. We are given that a positive integer \( N \) is divided into \( n \) parts inversely proportional to the sequence \( 2, 6, 12, 20, \ldots \). The smallest part is equal to \( \frac{1}{400} N \). We need to find the value of \( n \).

2. The sequence \( 2, 6, 12, 20, \ldots \) can be identified as the sequence of triangular numbers multiplied by 2. The \( k \)-th term of this sequence is given by:
   \[
   a_k = 2 \cdot \frac{k(k+1)}{2} = k(k+1)
   \]

3. Since the parts are inversely proportional to \( k(k+1) \), the parts can be written as:
   \[
   \frac{x}{1 \cdot 2}, \frac{x}{2 \cdot 3}, \frac{x}{3 \cdot 4}, \ldots, \frac{x}{n(n+1)}
   \]
   where \( x \) is a constant.

4. The sum of these parts is equal to \( N \):
   \[
   N = \sum_{k=1}^n \frac{x}{k(k+1)}
   \]

5. We know that:
   \[
   \sum_{k=1}^n \frac{1}{k(k+1)} = \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right)
   \]
   This is a telescoping series, and it simplifies to:
   \[
   \sum_{k=1}^n \left( \frac{1}{k} - \frac{1}{k+1} \right) = 1 - \frac{1}{n+1} = \frac{n}{n+1}
   \]

6. Therefore, we have:
   \[
   N = x \sum_{k=1}^n \frac{1}{k(k+1)} = x \cdot \frac{n}{n+1}
   \]

7. The problem states that the smallest part is \( \frac{N}{400} \). The smallest part corresponds to \( \frac{x}{n(n+1)} \):
   \[
   \frac{x}{n(n+1)} = \frac{N}{400}
   \]

8. Solving for \( x \):
   \[
   x = \frac{N \cdot n(n+1)}{400}
   \]

9. Substituting \( x \) back into the equation for \( N \):
   \[
   N = \left( \frac{N \cdot n(n+1)}{400} \right) \cdot \frac{n}{n+1}
   \]

10. Simplifying:
    \[
    N = \frac{N \cdot n^2}{400}
    \]

11. Dividing both sides by \( N \):
    \[
    1 = \frac{n^2}{400}
    \]

12. Solving for \( n \):
    \[
    n^2 = 400
    \]
    \[
    n = \sqrt{400}
    \]
    \[
    n = 20
    \]

The final answer is \( \boxed{20} \).