5. Given that $a$ and $b$ are positive integers, the fraction $\frac{a}{b}$, when converted to a decimal, contains the consecutive digits $\overline{2015}$. Then the minimum value of $b$ is $\qquad$ .
5. 129.

Given that multiplying a fraction by an appropriate power of 10 can result in the first four digits after the decimal point being $\overline{2015}$, we only need to find the smallest value of $b$ for the fraction $\frac{a}{b}=0.2015 \cdots$.
On one hand,
$\frac{a}{b}-\frac{1}{5}=0.2015 \cdots-0.2=0.0015$
$125 \text {. }$
$$

We will now test each case.
When $b=126$,
$\frac{25}{126}=0.1984 \cdots \neq 0.2015 \cdots$;
When $b=127$, $\frac{26}{127}=0.2047 \cdots$;
When $b=128$, $\frac{26}{128}=0.2031 \cdots$;
When $b=129$, $\frac{26}{129}=0.2015 \cdots$.
Therefore, the smallest value of $b$ is 129.