We have that \begin{align*} \frac{1}{a_n} &= \frac{1}{\sqrt{1 + \left( 1 + \frac{1}{n} \right)^2} + \sqrt{1 + \left( 1 - \frac{1}{n} \right)^2}} \\ &= \frac{\sqrt{1 + \left( 1 + \frac{1}{n} \right)^2} - \sqrt{1 + \left( 1 - \frac{1}{n} \right)^2}}{\left( \sqrt{1 + \left( 1 + \frac{1}{n} \right)^2} + \sqrt{1 + \left( 1 - \frac{1}{n} \right)^2} \right) \left( \sqrt{1 + \left( 1 + \frac{1}{n} \right)^2} - \sqrt{1 + \left( 1 - \frac{1}{n} \right)^2} \right)} \\ &= \frac{\sqrt{1 + \left( 1 + \frac{1}{n} \right)^2} - \sqrt{1 + \left( 1 - \frac{1}{n} \right)^2}}{1 + (1 + \frac{1}{n})^2 - 1 - (1 - \frac{1}{n})^2} \\ &= \frac{\sqrt{1 + \left( 1 + \frac{1}{n} \right)^2} - \sqrt{1 + \left( 1 - \frac{1}{n} \right)^2}}{\frac{4}{n}} \\ &= \frac{n \left( \sqrt{1 + \left( 1 + \frac{1}{n} \right)^2} - \sqrt{1 + \left( 1 - \frac{1}{n} \right)^2} \right)}{4} \\ &= \frac{\sqrt{n^2 + (n + 1)^2} - \sqrt{n^2 + (n - 1)^2}}{4}, \end{align*}so \[\frac{1}{a_n} = \frac{\sqrt{n^2 + (n + 1)^2} - \sqrt{(n - 1)^2 + n^2}}{4}.\]Hence, \begin{align*} &\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3} + \dots + \frac{1}{a_{100}} \\ &= \frac{\sqrt{1^2 + 2^2} - \sqrt{0^2 + 1^2}}{4} + \frac{\sqrt{2^2 + 3^2} - \sqrt{1^2 + 2^2}}{4} + \frac{\sqrt{3^2 + 4^2} - \sqrt{2^2 + 3^2}}{4} \\ &\quad + \dots + \frac{\sqrt{100^2 + 101^2} - \sqrt{99^2 + 100^2}}{4} \\ &= \boxed{\frac{\sqrt{20201} - 1}{4}}.\end{align*}