The triple $(f(x_1, x_2), A, B)$ does exist, so the answer is yes.
Let $A = O = (0, 0)$, $B = (x_0, y_0) = (2009 + \frac{2}{3}, \frac{1}{3})$. The idea is to search for a polynomial $f$ such that $f(x, y) = 0$ is the equation of an ellipse centered at $B$, passing through $O$ and with tangent line $y = 0$ at $O$. In fact, if $f$ is chosen like this, the ellipse $f(x, y) = 0$ is completely contained in the region $\{(x, y) ; 0 \le y \le \frac{2}{3}\}$, with $O$ the only integer point on the ellipse or in its interior; clearly, the absolute minimum of $f(x, y)$ is attained at $B$ and $f(x, y)$ is positive at all integer points other than $O$. Therefore, we consider polynomials of the type
$$f(X, Y) = 9M(X - x_0)^2 + 9N(X - x_0)(Y - y_0) + 9P(Y - y_0)^2 - Q$$
where $M, N, P, Q$ are integers with $M, P, Q, 4MP - N^2 > 0$.
The condition that the ellipse $f(x, y) = 0$ passes through $O$, with tangent line $y = 0$ at $O$, is expressed by
$$\begin{cases} 6029^2 M + 6029N + P - Q = 0 \\ 2 \cdot 6029M + N = 0. \end{cases}$$

It is then sufficient to choose $M = 1$, $N = -2 \cdot 6029$, $P$ any integer greater than $6029^2$ and $Q = P - 6029^2$.

(Alternative Solution. D. Schwarz)
Given any integer point $A(a_1, a_2)$, there exist infinitely many points $B(b_1, b_2)$ with $b_1, b_2 \in \mathbb{Q} \setminus \mathbb{Z}$, and such that $|a_1 - b_1| + |a_2 - b_2| = 2010$, for example $b_1 = a_1 + \alpha + r$, $b_2 = a_2 + \beta + (1 - r)$, with $\alpha, \beta \in \mathbb{Z}_+$, $r \in \mathbb{Q} \cap (0, 1)$, and $\alpha + \beta = 2009$. We now will consider polynomials of the type
$$f(X, Y) = N \left( (X - a_1)^2 + (Y - a_2)^2 + \varepsilon \right) \left( (X - b_1)^2 + (Y - b_2)^2 \right)$$
where $\varepsilon \in \mathbb{Q}_+^*$ and $N \in \mathbb{Z}_+^*$ large enough for $f(X, Y)$ to have integer coefficients.
One then has $f(b_1, b_2) = 0$, while $f(x, y) > 0$ for all points $(x, y)$ in the plane, other than $B$.
One also then has $f(a_1, a_2) = N\varepsilon((a_1 - b_1)^2 + (a_2 - b_2)^2)$, while one has, for all integer points $(n_1, n_2)$ in the plane, other than $A$,
$$\min \{(n_1 - a_1)^2 + (n_2 - a_2)^2 + \varepsilon\} = 1 + \varepsilon \quad \text{and} \\ \min \{(n_1 - b_1)^2 + (n_2 - b_2)^2\} = m,$$
for some $m \in \mathbb{Q} \cap (0, \frac{1}{2}]$ (for example $m = \frac{1}{2}$ when $r = \frac{1}{2}$), therefore $F(n_1, n_2) > N(1+\varepsilon)m$. In order to have $f(n_1, n_2) > f(a_1, a_2)$ it is thus enough that $N(1+\varepsilon)m \ge N\varepsilon((a_1 - b_1)^2 + (a_2 - b_2)^2)$, therefore let us take
$$\varepsilon = m / ((a_1 - b_1)^2 + (a_2 - b_2)^2 - m) \in \mathbb{Q}_+^*,$$
and then choose some appropriate $N$.