Divide 15 into two parts such that sum of reciprocals is $\dfrac{3}{10}.$

Let numbers be x and (15 - x).

According to the question,

$\Rightarrow \dfrac{1}{x} + \dfrac{1}{15 - x} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{15 - x + x}{x(15 - x)} = \dfrac{3}{10} \\[1em] \Rightarrow \dfrac{15}{x(15 - x)} = \dfrac{3}{10} \\[1em] \Rightarrow 150 = 3x(15 - x) \\[1em] \Rightarrow 150 = 45x - 3x^2 \\[1em] \Rightarrow 3x^2 - 45x + 150 = 0 \\[1em] \Rightarrow 3(x^2 - 15x + 50) = 0 \\[1em] \Rightarrow x^2 - 10x - 5x + 50 = 0 \\[1em] \Rightarrow x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (x - 5)(x - 10) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow x = 5 \text{ or } x = 10.$

Hence, numbers = 5, 10.