1. We start with the given equation:
$$\frac{1}{d_{1}} + \frac{1}{d_{2}} + \cdots + \frac{1}{d_{k}} = \frac{1620}{1003}$$
where $d_1, d_2, \ldots, d_k$ are all different divisors of $n$.

2. Notice that $1003 = 17 \times 59$. This implies that for the sum of the reciprocals of the divisors to have a denominator of $1003$, $n$ must be divisible by both $17$ and $59$. Since $n$ is even, $n$ must also be divisible by $2$. Therefore, $n$ must be divisible by $2 \times 17 \times 59 = 2006$.

3. Next, we need to verify that the sum of the reciprocals of all divisors of $2006$ equals $\frac{1620}{1003}$. The divisors of $2006$ are $1, 2, 17, 34, 59, 118, 1003, 2006$.

4. We calculate the sum of the reciprocals of these divisors:
$$\sum_{d|2006} \frac{1}{d} = \frac{1}{1} + \frac{1}{2} + \frac{1}{17} + \frac{1}{34} + \frac{1}{59} + \frac{1}{118} + \frac{1}{1003} + \frac{1}{2006}$$

5. To find a common denominator, we use $2006 = 2 \times 17 \times 59$:
$$\frac{1}{1} = \frac{2006}{2006}, \quad \frac{1}{2} = \frac{1003}{2006}, \quad \frac{1}{17} = \frac{118}{2006}, \quad \frac{1}{34} = \frac{59}{2006}$$
$$\frac{1}{59} = \frac{34}{2006}, \quad \frac{1}{118} = \frac{17}{2006}, \quad \frac{1}{1003} = \frac{2}{2006}, \quad \frac{1}{2006} = \frac{1}{2006}$$

6. Summing these fractions:
$$\frac{2006 + 1003 + 118 + 59 + 34 + 17 + 2 + 1}{2006} = \frac{3240}{2006}$$

7. Simplifying $\frac{3240}{2006}$:
$$\frac{3240}{2006} = \frac{3240 \div 2}{2006 \div 2} = \frac{1620}{1003}$$

8. Therefore, the sum of the reciprocals of the divisors of $2006$ is indeed $\frac{1620}{1003}$.

9. Since $n$ must be a multiple of $2006$ to satisfy the given equation, all solutions for $n$ are multiples of $2006$.

The final answer is $\boxed{ n = 2006k }$ for $k \in \mathbb{N}$.