Your mind will explode 3

First of all, suppose $251$ appears in the expansion of $m/n$ . Then $10^k m/n = 0.251\ldots$ plus an integer $d$ , so $\dfrac{10^k m - nd}{n} = 0.251 \ldots$ .

So we can assume that $0.251 \le m/n < 0.252$ . Since $n < 4m$ , we can write $n = 4m-a$ for some positive integer $a$ . Then we have $\dfrac{m}{4m-a} < 0.252,$ which leads after some algebra to $m > 63a/2$ . Since we are trying to minimize $n$ , we should clearly take $a = 1$ (otherwise $m$ would be at least $63$ and $n$ would be at least $63/0.252 = 250$ ). So the smallest $m$ we can take is $m = 32$ , $n = \fbox{127}$ .

To find the smallest value of $n,$ we consider when the first three digits after the decimal point are $0.251\ldots$ .

Otherwise, suppose the number is in the form of $\dfrac{m}{n} = 0.X251 \ldots,$ were $X$ is a string of $k$ digits and $n$ is small as possible. Then $10^k \cdot \dfrac{m}{n} - X = \dfrac{10^k m - nX}{n} = 0.251 \ldots$ . Since $10^k m - nX$ is an integer and $\dfrac{10^k m - nX}{n}$ is a fraction between $0$ and $1,$ we can rewrite this as $\dfrac{10^k m - nX}{n} = \dfrac{p}{q},$ where $q \le n.$ Then the fraction $\dfrac pq = 0.251 \ldots$ suffices.

So we have $\dfrac{m}{n} = 0.251\ldots$ , or

$\frac{251}{1000} \le \frac{m}{n} < \frac{252}{1000} \Longleftrightarrow 251n \le 1000m < 252n \Longleftrightarrow n \le 250(4m-n) < 2n.$

As $4m > n,$ we know that the minimum value of $4m - n$ is $1;$

hence we need $250 < 2n \Longrightarrow 125 < n$ . Since $4m - n = 1,$ we need $n + 1$ to be divisible by $4,$ and this first occurs when $n = \boxed{ 127 }$