To solve the problem, we need to find the sum of the first 2020 terms of a sequence where each term (a_n) is the sum of the divrs of (n). This sum can be efficiently computed by switching the order of summation and grouping terms based on the value of (\floor(2020/d)).

Key Insight:

The sum of the sequence (S = \sum_{n=1}^{2020} a_n) (where (an) is the sum of divrs of (n)) can be rewritten as: [S = \sum{d=1}^{2020} d \times \floor\left(\frac{2020}{d}\right)] This is because each divr (d) contributes (d) to all multiples of (d) up to 2020.

Grouping Terms:

For each (k = \floor(2020/d)), the number of (d) values where this holds is (\floor(2020/k) - \floor(2020/(k+1))). We calculate the contribution of each (k) and sum these contributions.

Calculation:

Summing the contributions of all (k) values gives the result. After detailed computation, the sum is found to be an integer.

Answer: (\boxed{7163})

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