Show it suffices for $251$ to appear immediately after the decimal point, so $\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000}$

Use the neighboring fractions $\frac14 \lt \frac{32}{127} \lt \frac{63}{250}:$ each adjacent pair has cross-difference $1,$ which bounds the denominators between them

It suffices to make $251$ appear immediately after the decimal point: if $\frac{m}{n} = 0.A251\ldots$ with $A$ a block of $k \ge 1$ digits, then $10^k \frac{m}{n} - A = 0.251\ldots$ is a fraction between $0$ and $1$ whose reduced denominator is at most $n.$ So we need the smallest $n$ admitting an $m$ with $$\frac{251}{1000} \le \frac{m}{n} \lt \frac{252}{1000},$$ that is $$0 \le 1000m - 251n \lt n.$$

The fraction $\frac{32}{127}$ lies in this interval because $$\frac{251}{1000} \lt \frac{32}{127} \lt \frac{63}{250} = \frac{252}{1000}.$$ It remains to prove that no smaller denominator works. We use the following elementary fact: if $\frac{a}{b} \lt \frac{u}{v} \lt \frac{c}{d}$ and $bc-ad=1,$ then $$\begin{aligned} v &= b(cv-du)+d(bu-av) \\ &\ge b+d, \end{aligned}$$ because both parenthesized quantities are positive integers.

Now $4\cdot32-1\cdot127=1$ and $127\cdot63-32\cdot250=1.$ Therefore every fraction strictly between $\frac14$ and $\frac{32}{127}$ has denominator at least $4+127=131,$ while every fraction strictly between $\frac{32}{127}$ and $\frac{63}{250}$ has denominator at least $127+250=377.$ Since our target interval lies inside $\left(\frac14,\frac{63}{250}\right)$ and contains $\frac{32}{127},$ no fraction in it has denominator below $127.$

The smallest possible value of $n$ is $127.$