3. Let the function $f: \mathbf{R} \rightarrow \mathbf{R}$, for any $x, y \in \mathbf{R}$, we have $f(x+y)=f(x)+f(y)$, and when $x<0$, $f(x)<0$. If the graph of $y=f(x)$ intersects or is tangent to the line $2007 x + 2008 y = 2009$, then the number of intersection or tangent points is ( ).
(A) 1
(B) 2
(C) 3
(D) Uncertain, depends on the specific form of $y=f(x)$
3. A.

From the problem, for any $x, y \in \mathbf{R}$, we have
$$
f(x+y)=f(x)+f(y) \text {. }
$$

Taking $x=y=0$ in equation (1), we get
$$
f(0)=0 \text {. }
$$

Taking $y=-x$ in equation (1) and using equation (2), we get
$$
f(-x)=-f(x) \text {. }
$$

According to the problem and equation (3), when $x>0$,
$$
f(x)=-f(-x)>0 \text {. }
$$

For any $x \in \mathbf{R}, t>0$, using equations (1) and (4), we get
$$
f(x+t)=f(x)+f(t)>f(x) \text {. }
$$

Therefore, $f$ is an increasing function on $\mathbf{R}$. Thus, for $x_{1}>x_{2}$, we have
$$
2007 x_{1}+2008 f\left(x_{1}\right)>2007 x_{2}+2008 f\left(x_{2}\right) \text {. }
$$

Hence, $2007 x+2008 f(x)$ is an increasing function.
Therefore, there is at most one $x_{1} \in \mathbf{R}$ such that
$$
2007 x_{1}+2008 f\left(x_{1}\right)=2009
$$
(From the problem, such an $x_{1}$ exists).
Thus, the graph of $y=f(x)$ intersects the line
$$
2007 x+2008 y=2009
$$

at exactly one point or is tangent to it.