Question 211, Find the largest integer $k$, such that $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{\mathrm{n}^{2}}{\sqrt{3 \mathrm{n}^{2}-k}}$ holds for all positive integers $n \geq 2$.

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Question 211, Solution Method One: We first test the first few values.
When $\mathrm{n}=2$, we require $2>\frac{4}{\sqrt{12-k}}$, which gives $\mathrm{k} \leq 7$;
When $n=3$, we require $2>\frac{9}{\sqrt{27-k}}$, which gives $k \leq 6$;
When $n=4$, we require $3>\frac{16}{\sqrt{48-k}}$, which gives $k \leq 5$;
When $n=5$, we require $3>\frac{25}{\sqrt{75-k}}$, which gives $k \leq 6$;
When $n=6$, we require $4>\frac{36}{\sqrt{108-k}}$, which gives $k \leq 26$;
Thus, $k \leq 5$. Next, we prove that $k=5$ satisfies the condition, which means we only need to prove that for $n \geq 7$, we always have $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{\mathrm{n}^{2}}{\sqrt{3 \mathrm{n}^{2}-5}}$.

Let $\left[\frac{n}{\sqrt{3}}\right]+1=m$, then $m>\frac{n}{\sqrt{3}}$, i.e., $3 m^{2}-n^{2}>0$. Since $3 m^{2}-n^{2} \equiv 2,3(\bmod 3)$, we have $3 m^{2}-n^{2} \geq 2$, thus $\left[\frac{n}{\sqrt{3}}\right]+1=m \geq \sqrt{\frac{n^{2}+2}{3}}>\frac{n^{2}}{\sqrt{3 n^{2}-5}}$.
In summary, the maximum value of the positive integer $\mathrm{k}$ is 5.
Solution Method Two: Let $\left[\frac{n}{\sqrt{3}}\right]+1=m$, then $m>\frac{n}{\sqrt{3}}$, i.e., $3 m^{2}-n^{2}>0$. Since $3 m^{2}-n^{2} \equiv 2,3(\bmod 3)$, we have $3 m^{2}-n^{2} \geq 2$. On the other hand, the equation $3 y^{2}-x^{2}=2$ has a solution $(x, y)=(1,1)$, hence there are infinitely many positive integer solutions. Therefore, there are infinitely many positive integers $n$ such that $\left[\frac{n}{\sqrt{3}}\right]+1=m=\sqrt{\frac{n^{2}+2}{3}}$.
Since $\sqrt{\frac{n^{2}+2}{3}}>\frac{n^{2}}{\sqrt{3 n^{2}-k}} \Leftrightarrow(6-k) n^{2}-2 k>0$, we have $k \leq 5$.
When $n=2$, we verify that $k \leq 7$; when $n \leq 3$, we verify that $k \leq 6$. Therefore, for any positive integer $n \geq 2$, we have $\left[\frac{n}{\sqrt{3}}\right]+1>\frac{n^{2}}{\sqrt{3 n^{2}-5}}$.
In summary, the maximum value of the positive integer $\mathrm{k}$ is 5.