Answer: 750

Small Hint:

Convert the repeating three-digit block into a fraction with denominator $999$

Big Hint:

Use the integrality of $n$ to determine which digit makes $4d+1$ divisible by $37$

Solution:

The repeating decimal is $\frac{100d+25}{999}.$ Thus $$\begin{aligned}n&=\frac{810(100d+25)}{999}\\&=\frac{750(4d+1)}{37}.\end{aligned}$$ Since $0\leq d\leq9,$ the only way $4d+1$ can be divisible by $37$ is $4d+1=37,$ so $d=9$ and $n=750.$