For all \(n \in \mathbb{N}\), let \(\tau (n)\) denote the sum of positive divisors of \(n\) (including \(1\) and itself). Find the last three digits of \(\displaystyle \sum \limits_{n=1}^{100} \tau (n) \).

**Details and assumptions**

As an explicit example, the positive divisors of \(4\) are \(1, 2\) and \(4\), so \(\tau (4)= 4+2+1 = 7\).

This is

**not**a computer science problem.

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