To solve the problem, we first determine the number of positive divrs of 2020.

Step 1: Calculate the number of divrs of 2020

Factorize 2020:
[2020 = 2^2 \times 5^1 \times 101^1]
The number of divrs is ((2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12).

Step 2: Interpret the problem

The problem asks for the sum of positive integers (n) where the number of divrs of (n) equals 12. However, without a constraint (like (n \leq 2020)), the sum is infinite. Assuming the intended constraint is (n = 2020) (since it's the only multiple of itself with 12 divrs), the sum is 2020.

Answer: (\boxed{2020})

(Note: If the constraint was (n \leq 100), the sum is 402, but 2020 is the most plausible intended answer.)

(\boxed{2020})

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