3. Let $n$ be a positive integer, such that $\sqrt{3}$ lies between $\frac{n+3}{n}$ and $\frac{n+4}{n+1}$. Then $n=$ $\qquad$ .
3.4.
Notice that,
$$
\frac{n+4}{n+1}=1+\frac{3}{n+1}<1+\frac{3}{n}=\frac{n+3}{n} \text{.}
$$

Since $\sqrt{3}$ is an irrational number, then
$$
\begin{array}{l}
1+\frac{3}{n+1}<\sqrt{3}<1+\frac{3}{n} \\
\Rightarrow \frac{3}{n+1}<\sqrt{3}-1<\frac{3}{n} \\
\Rightarrow n<\frac{3}{\sqrt{3}-1}<n+1 .
\end{array}
$$

Thus, $n=\left[\frac{3}{\sqrt{3}-1}\right]=4$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).