3. The quadratic equation in $x$
$$
6 x^{2}-(2 m-1) x-(m+1)=0
$$

has a root $\alpha$, given that $\alpha$ satisfies $|\alpha| \leqslant 2000$, and makes $\frac{3}{5} \alpha$ an integer. Then the number of possible values for $m$ is $\qquad$.
3.2 401 .

From the quadratic formula, we get
$$
x_{1,2}=\frac{(2 m-1) \pm(2 m+5)}{12},
$$

which gives the roots as $-\frac{1}{2}$ and $\frac{m+1}{3}$.
If $\alpha=-\frac{1}{2}$, then $\frac{3}{5} \alpha=-\frac{3}{10}$ is not an integer. Therefore,
$$
\alpha=\frac{m+1}{3} \text {. }
$$

Given $|\alpha| \leqslant 2000$, we have $-2000 \leqslant \frac{m+1}{3} \leqslant 2000$, which implies $-6000 \leqslant m+1 \leqslant 6000$, and $\alpha$ must make $\frac{3}{5} \alpha=\frac{m+1}{5}$ an integer. Thus,
$$
\begin{array}{l}
m=-1+5 n(n=0, \pm 1, \pm 2, \cdots) . \\
\text { Hence }-6000 \leqslant(-1+5 n)+1 \leqslant 6000 .
\end{array}
$$

Solving this, we get $-1200 \leqslant n \leqslant 1200$.
Thus, $n=0, \pm 1, \pm 2, \cdots, \pm 1200$.
Therefore, the number of possible values for $m$ is $1200 \times 2+1=2401$.