3. $m, n$ are positive integers. If $\frac{2000}{2001}<\frac{n}{m}<\frac{2001}{2002}$, then the fraction $\frac{n}{m}=$ $\qquad$ when $m$ is the smallest.
3. $\frac{4001}{4003}$.

First, give the general solution formula: $m, n, a, b, c, d$ are all positive integers, and $\frac{a}{b}m a, m c>n d$. We can set
$\left\{\begin{array}{l}b n-a m=x, \\ c m-d n=y .\end{array}(x, y\right.$ are positive integers $)$
Solving, we get $\left\{\begin{array}{l}m=\frac{d x+b y}{b c-a d}, \\ n=\frac{c x+a y}{b c-a d} .\end{array}\right.$
$$
\begin{array}{l}
\because b c-a d=2001^{2}-2000 \times 2002=1, \\
\therefore\left\{\begin{array}{l}
m=2002 x+2001 y, \\
n=2001 x+2000 y .
\end{array}\right.
\end{array}
$$

When $x=y=1$, $m=4003, n=4001$.
Since $m$ and $n$ are coprime, $\frac{n}{m}=\frac{4001}{4003}$.