276. In the decimal representation of a positive number $a$, all decimal places starting from the fourth digit after the decimal point are discarded, i.e., the number $a$ is rounded down to $a_{0}$ with an accuracy of 0.001. The obtained number $a_{0}$ is divided by the original number $a$, and the quotient is again rounded down with the same accuracy. List all the numbers that can result from this process.

$277^{*}$. Let $\alpha$ be an arbitrary irrational number. Clearly, for any integer $n$, the fraction in the series $\frac{0}{n}=0, \frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \ldots$ that is closest to $\alpha$ differs from $\alpha$ by no more than half of $\frac{1}{n}$. Prove that there exist such $n$ for which the fraction with denominator $n$, closest to $\alpha$, differs from $\alpha$ by less than $0.001 \cdot \frac{1}{n}$.
276. It is clear that for $a\frac{1}{2} \quad(\text { and } d \leqslant 1)
$$

- this estimate of the fraction $d$ is already final, because the ratio $\delta=\frac{\alpha}{a}$ can take any value within the range $0 \leqslant \delta < \frac{1}{2}$ (which means $d$ can take any value within the range $\frac{1}{2} < d \leqslant 1$). Indeed, by setting $a_{0}$ equal to $0.001$, we get that the fraction $\delta=$

$$
\begin{array}{ll}
=\frac{\alpha}{a_{0}+\alpha}=\frac{\alpha}{0.001+\alpha}, & \text { from which } \\
& \alpha=0.001 \frac{\delta}{1-\delta}
\end{array}
$$

such $\alpha$ for any $\delta$ (where $0 \leqslant \delta < \frac{1}{2}$) satisfies the inequality $0 \leqslant \alpha < 0.001$, and the corresponding $a=0.001+\alpha$ corresponds to the value $d=1-\delta$. It is clear that if $\delta$ takes all values between 0 and $\frac{1}{2}$, then $d$ takes all values from $\frac{1}{2}$ to 1, i.e., $d$ can have any of the following values: $0; 0.5; 0.501; 0.502; \ldots; 0.999; 1$.