jmc

To find the smallest value of $n$, we consider when the first three digits after the decimal point are $0.251\dots$. Otherwise, suppose the number is in the form of $\frac{m}{n}=0.X251\dots$, where $X$ is a string of $k$ digits and $n$ is small as possible. Then $10^k\cdot \frac{m}{n}-X=\frac{10^{k}m-nX}{n}=0.251\dots$. Since $10^{k}m-nX$ is an integer and $\frac{10^{k}m-nX}{n}$ is a fraction between $0$ and $1$, we can rewrite this as $\frac{10^{k}m-nX}{n}=\frac{p}{q}$, where $q\le n$. Then the fraction $\frac{p}{q}=0.251\dots$ suffices. Thus we have $\frac{m}{n}=0.251\dots$, or $\frac{251}{1000}\le \frac{m}{n}<\frac{252}{1000}⟺251n\le 1000m<252n⟺n\le 250(4m-n)<2n.$ As $4m>n$, we know that the minimum value of $4m-n$ is $1$; hence we need $250<2n⟹125<n$. Since $4m-n=1$, we need $n+1$ to be divisible by $4$, and this first occurs when $n=\boxed{127}$.