3. Given that $n$ is a natural number, $9 n^{2}-10 n+2009$ can be expressed as the product of two consecutive natural numbers. Then the maximum value of $n$ is $\qquad$

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3.2007 .

Let $9 n^{2}-10 n+2009=m(m+1)$, where $m$ is a natural number. Then
$$
9 n^{2}-10 n+\left(2009-m^{2}-m\right)=0 \text {. }
$$

Considering equation (1) as a quadratic equation in the natural number $n$, its discriminant should be the square of a natural number. Let's assume $\Delta=t^{2}(t \in \mathbf{N})$, then
$$
(-10)^{2}-4 \times 9\left(2009-m^{2}-m\right)=t^{2} \text {. }
$$

Simplifying and rearranging, we get
$$
(6 m+3)^{2}-t^{2}=72233,
$$

which is $(6 m+3+t)(6 m+3-t)=72233$.
Let $6 m+3+t=a, 6 m+3-t=b$. Then $t=\frac{1}{2}(a-b)$.
When $a=72233, b=1$, $t$ has its maximum value.
At this point, the maximum value of $t$ is 36116.
$$
\text { Also, } n=\frac{-(-10) \pm \sqrt{\Delta}}{2 \times 9}=\frac{10 \pm \sqrt{t^{2}}}{18}=\frac{10 \pm t}{18} \text {, }
$$

When $t$ takes its maximum value of 36116, $n$ takes its maximum value, which is
$$
\frac{10+36116}{18}=2007 .
$$