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If $$\sum_{r=2}^{100} \frac{1}{1 + 2 + 3 + \cdots + r} = \frac{x}{y}$$, where $$x$$, $$y$$ are positive integers, GCD of $$x$$, $$y$$ is 1, then $$x + y =$$
$$\frac{1}{1 + 2 + \cdots + r} = \frac{2}{r(r + 1)} = 2\left(\frac{1}{r} - \frac{1}{r + 1}\right)$$. The sum telescopes to $$2\left(\frac{1}{2} - \frac{1}{101}\right) = \frac{99}{101}$$, which is in lowest terms. So $$x + y = 99 + 101 = 200$$.
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