Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$
Denote $x = \log_b n$.
Hence, the system of equations given in the problem can be rewritten as
\begin{align*} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{align*}
Solving the system gives $x = 4$ and $b = \frac{5}{4}$.
Therefore,
\[n = b^x = \frac{625}{256}.\]
Therefore, the answer is $625 + 256 = \boxed{881}$.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)