It isn't 2013!

For \(1\le i\le 10000\), an integer is chosen uniformly at random between 1 and \(i\) inclusive. Let \(S\) be the sum of all numbers chosen in this way. If the probability that \(2014\) divides \(S\) but \(2013\) does not divide \(S\) can be expressed as \(\frac{a}{b}\), where \(a\) and \(b\) are relatively prime positive integers, find the remainder when \(a+b\) is divided by 1000.

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