4. If the positive integer $n$ makes $\sqrt{3}$ always lie between $1+\frac{3}{n}$ and $1+\frac{3}{n+1}$, then $n=$ . $\qquad$
4. 4 .

When $n$ is a positive integer, it is easy to know that $\frac{3}{n+1}<\frac{3}{n}$.
$$
\begin{array}{l}
\text { By } \frac{3}{n+1}<\sqrt{3}-1<\frac{3}{n} \\
\Rightarrow \frac{1+3 \sqrt{3}}{2}<n<\frac{3}{\sqrt{3}-1}=\frac{3+3 \sqrt{3}}{2} .
\end{array}
$$

Therefore, $n \in\left(\frac{1+3 \sqrt{3}}{2}, \frac{3+3 \sqrt{3}}{2}\right)$, this interval has a length of 1, and contains only the unique integer 4.
Thus, $n=4$.