4. If the positive integer $n$ makes $\sqrt{3}$ always lie between $1+\frac{3}{n}$ and $1+\frac{3}{n+1}$, then $n=$
Answer: 4.
Solution: When $n$ is a positive integer, it is easy to know that $\frac{3}{n+1}<\frac{3}{n+1}$. We get $n>\frac{1+3 \sqrt{3}}{2}$. Therefore, $n \in\left(\frac{1+3 \sqrt{3}}{2}, \frac{3+3 \sqrt{3}}{2}\right)$. The length of this interval is 1, and it contains only one integer, 4. So $n=4$.