1. **Problem Statement:** Determine the largest positive integer $n$ for which there exist positive integers $a$ and $q$ such that $$q^6 \leq n \quad \text{and} \quad \left|\sqrt{2} - \frac{a}{q}\right| \leq \frac{1}{\sqrt{n}}.$$ 2. **Understanding the problem:** We want to find the maximum $n$ such that there is a rational approximation $\frac{a}{q}$ to $\sqrt{2}$ with denominator $q$ satisfying the two inequalities. 3. **Rewrite the inequalities:** From $q^6 \leq n$, we have $n \geq q^6$. From $\left|\sqrt{2} - \frac{a}{q}\right| \leq \frac{1}{\sqrt{n}}$, substituting $n \geq q^6$ gives $$\left|\sqrt{2} - \frac{a}{q}\right| \leq \frac{1}{\sqrt{n}} \leq \frac{1}{\sqrt{q^6}} = \frac{1}{q^3}.$$ 4. **Key inequality:** We need positive integers $a,q$ such that $$\left|\sqrt{2} - \frac{a}{q}\right| \leq \frac{1}{q^3}.$$ 5. **Approximation of irrationals by rationals:** By Dirichlet's approximation theorem, for any irrational number $\alpha$, there exist infinitely many rationals $\frac{a}{q}$ such that $$\left|\alpha - \frac{a}{q}\right| < \frac{1}{q^2}.$$ However, here the bound is stronger: $\frac{1}{q^3}$. 6. **Known results:** For quadratic irrationals like $\sqrt{2}$, the best possible approximation exponent is 2, meaning $$\left|\sqrt{2} - \frac{a}{q}\right| \geq \frac{c}{q^2}$$ for some positive constant $c$ and all $a,q$. 7. **Conclusion:** Since $\frac{1}{q^3}$ is smaller than $\frac{c}{q^2}$ for large $q$, the inequality $$\left|\sqrt{2} - \frac{a}{q}\right| \leq \frac{1}{q^3}$$ cannot hold for arbitrarily large $q$. 8. **Check small values of $q$:** Try $q=1$: $$\left|\sqrt{2} - a\right| \leq 1,$$ choose $a=1$, $|1.4142 -1|=0.4142 \leq 1$ true. Check $q=2$: $$\left|\sqrt{2} - \frac{a}{2}\right| \leq \frac{1}{8} = 0.125.$$ Try $a=3$, $|1.4142 - 1.5|=0.0858 \leq 0.125$ true. Check $q=3$: $$\left|\sqrt{2} - \frac{a}{3}\right| \leq \frac{1}{27} \approx 0.037.$$ Try $a=4$, $|1.4142 - 1.3333|=0.0809 > 0.037$ no. Try $a=5$, $|1.4142 - 1.6667|=0.2525 > 0.037$ no. No $a$ satisfies for $q=3$. 9. **Therefore, the largest $q$ for which the inequality holds is $q=2$. 10. **Find $n$:** Since $n \geq q^6$, for $q=2$, $n \geq 2^6 = 64$. 11. **Check if $n=64$ works:** $$q^6 = 64 \leq 64 = n,$$ $$\left|\sqrt{2} - \frac{3}{2}\right| = 0.0858 \leq \frac{1}{\sqrt{64}} = \frac{1}{8} = 0.125,$$ which is true. 12. **Check if $n > 64$ can work:** For $n > 64$, $q^6 \leq n$ implies $q \geq 3$ for $n \geq 729$. But for $q=3$, no $a$ satisfies the inequality. 13. **Final answer:** The largest positive integer $n$ is $$\boxed{64}.$$