Solution:

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

Thus $251n$ must land within $n$ below a multiple of $1000.$ Try $n = 4m - 1:$ then $251n = 251(4m - 1) = 1000m + (4m - 251),$ so for $m \le 62$ this lies below $1000m$ by $251 - 4m.$ The requirement $251 - 4m \lt n = 4m - 1$ gives $m \gt 31.5,$ so $m = 32$ and $n = 127$ work: indeed $\frac{32}{127} = .2519\ldots$ A short check of the same inequality shows no smaller $n$ puts $1000m$ within $n$ above a multiple of $251,$ since the deficit $251 - 4m$ (or its analogues for other residues) stays too large.

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