For any positive real number $x, \lfloor x \rfloor$ equals the largest integer less than or equal to $x$ and so $\lfloor x \rfloor \leq x$. In particular, $\lfloor x \rfloor \cdot x \leq x \cdot x = x^{2}$. Thus, if $\lfloor x \rfloor \cdot x = 36$, then $36 \leq x^{2}$. Since $x > 0$, then $x \geq 6$. In fact, if $x = 6$, then $\lfloor x \rfloor = \lfloor 6 \rfloor = 6$ and so $\lfloor x \rfloor \cdot x = x^{2} = 36$. Therefore, $x = 6$. (Note that if $x > 6$, then $\lfloor x \rfloor \cdot x > 6 \cdot 6 = 36$.) Also, since $\lfloor y \rfloor \cdot y = 71$, then $y^{2} \geq 71$. Since $y > 0$, then $y \geq \sqrt{71} \approx 8.43$. Since $y \geq \sqrt{71} \approx 8.43$, then $\lfloor y \rfloor \geq 8$. Suppose that $\lfloor y \rfloor = 8$. In this case, $y = \frac{71}{\lfloor y \rfloor} = \frac{71}{8} = 8.875$. Note that if $y = \frac{71}{8}$, then $\lfloor y \rfloor = 8$, so $y = \frac{71}{8}$ is a solution. Therefore, $x + y = 6 + \frac{71}{8} = \frac{119}{8}$.