Solution:

Among fractions with numerator $1$ or $2,$ the closest neighbors of $\frac{2}{7} \approx 0.2857$ are $\frac{1}{4} = 0.25$ below (note $\frac{2}{8} = \frac{1}{4}$) and $\frac{1}{3} \approx 0.3333$ above (note $\frac{2}{6} = \frac{1}{3}$); no other candidate lies between them. So $\frac{2}{7}$ is the unique closest fraction to $r$ exactly when $r$ is closer to $\frac{2}{7}$ than to both $\frac{1}{4}$ and $\frac{1}{3},$ i.e. when $r$ lies strictly between the midpoints $$\frac{1}{2}\left(\frac{1}{4} + \frac{2}{7}\right) = \frac{15}{56} = 0.26785\ldots \qquad \text{and} \qquad \frac{1}{2}\left(\frac{2}{7} + \frac{1}{3}\right) = \frac{13}{42} = 0.30952\ldots.$$

The four-place decimals in that interval are $0.2679, 0.2680, \ldots, 0.3095,$ and there are $3095 - 2679 + 1 = 417$ of them.