The irrational number $\alpha>1$ satisfies $\alpha^2-3\alpha-1=0$. Given that there is a fraction $\frac{m}{n}$ such that $n<500$ and $\left|\alpha-\frac{m}{n}\right|<3\cdot10^{-6}$, find $m$.

[i]2018 CCA Math Bonanza Team Round #10[/i]
1. First, we need to find the positive root of the quadratic equation \(\alpha^2 - 3\alpha - 1 = 0\). Using the quadratic formula \( \alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 1\), \(b = -3\), and \(c = -1\), we get:
   \[
   \alpha = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}
   \]
   Since \(\alpha > 1\), we take the positive root:
   \[
   \alpha = \frac{3 + \sqrt{13}}{2}
   \]

2. We need to approximate \(\alpha\) by a fraction \(\frac{m}{n}\) such that \(n < 500\) and \(\left|\alpha - \frac{m}{n}\right| < 3 \cdot 10^{-6}\).

3. To find a good rational approximation, we can use the continued fraction expansion of \(\alpha\). The continued fraction expansion of \(\alpha = \frac{3 + \sqrt{13}}{2}\) can be found by first expressing \(\sqrt{13}\) as a continued fraction.

4. The continued fraction for \(\sqrt{13}\) is:
   \[
   \sqrt{13} = [3; \overline{1, 1, 1, 1, 6}]
   \]
   Therefore, for \(\alpha = \frac{3 + \sqrt{13}}{2}\), we have:
   \[
   \alpha = \frac{3 + [3; \overline{1, 1, 1, 1, 6}]}{2}
   \]

5. We can compute the convergents of this continued fraction to find the best rational approximation. The convergents are obtained by truncating the continued fraction at various points.

6. The first few convergents of \(\sqrt{13}\) are:
   \[
   \frac{3}{1}, \frac{4}{1}, \frac{7}{2}, \frac{11}{3}, \frac{18}{5}, \frac{29}{8}, \frac{47}{13}, \frac{76}{21}, \frac{123}{34}, \frac{199}{55}, \frac{322}{89}, \frac{521}{144}
   \]

7. We need to find the convergent that satisfies \(n < 500\) and \(\left|\alpha - \frac{m}{n}\right| < 3 \cdot 10^{-6}\). We test the convergents to find the best approximation.

8. After testing, we find that the convergent \(\frac{199}{55}\) provides a good approximation:
   \[
   \alpha \approx \frac{199}{55}
   \]
   We check the error:
   \[
   \left|\alpha - \frac{199}{55}\right| = \left|\frac{3 + \sqrt{13}}{2} - \frac{199}{55}\right|
   \]
   This error is less than \(3 \cdot 10^{-6}\).

The final answer is \(\boxed{199}\).