Solution:

Pulling the exponents out of the logarithms, the conditions become $$x \log_x y = 10 \qquad \text{and} \qquad 4y \log_y x = 10.$$ Multiplying these equations and using $\log_x y \cdot \log_y x = 1$ gives $4xy = 100,$ so $xy = 25.$

Such $x$ and $y$ do exist: the system solves to $\log_x y = \frac{10}{x}$ with $y = \frac{25}{x},$ which has a solution with $x, y \gt 1,$ so the answer is $25.$