Let the two consecutive integers be $n$ and $(n + 1)$. We are given that $\dfrac{1}{n} + \dfrac{1}{n + 1} = \dfrac{11}{30}$. Multiplying both sides by $30n(n + 1)$ gives $30(n + 1) + 30n = 11(n)(n + 1)$. Expanding the right side gives $30n + 30 + 30n = 11n^2 + 11n$. Combining like terms gives $60n + 30 = 11n^2 + 11n$. Moving all of the terms to the left side gives $11n^2 - 49n - 30 = 0$. Factoring gives $(n - 6)(11n + 5) = 0$. The possible solutions are $n = 6$ and $n = -\dfrac{5}{11}$. Since the second solution is not an integer, we have $n = \boxed{6}$.
