Can anybody please provide solutions or ideas?

Here's some ideas on how you might go about finding a solution. Let's say that for example we're trying to take $B=(1005.5,1004.5)$, $A=(0,0)$. Our goal will be to construct a polynomial such that (1) $f(x,y) \geq 0$ everywhere. (2). $f(1005.5,1004.5)=0$. Nowhere else is $f$ equal to $0$. (3). $f(0,0)$, while nonzero, is still the smallest value taken on by $f$ on an integer. We can satisfy (1) by making sure we right $f(x,y)$ as a product of polynomials which are either squares or sum of squares. So we just need to choose the factors right. To satisfy (2): Find a polynomial which is $0$ only at $(1005.5, 1004.5)$, and nowhere else. As long as this is one of your factors, and none of your factors have real roots, you should be good. To satisfy (3): You want to look a polynomial which does not itself have real roots, but is HUGE at every integer except for $(0,0)$ (HUGE in this case means that $f(0,0)$ is so much smaller than the other values that it dwarfs any contribution from the factor you found in the previous step). Once you find the two polynomials meeting the criteria above, take their product and you'll have one that meets all the original criteria.

here is may solution let $A(0,0) B(\frac{1}{4},\frac{8039}{4})$ and $f(x_1,x_2)=(x_1-\frac{1}{4})^{2}+(8039x_1-x_2)^2+\frac{3}{4}$ and we can easily prove that this function is suitable

A slight correction to the above post: You can use the same idea, but a better polynomial to use is $(4x_1-1)^2+(8039x_1-x_2)^2$. $A = (0, 0)$ and $B = (\frac{1}{4}, \frac{8039}{4})$ still. Firstly, it's obvious that this is an integer polynomial. Secondly, $f$ attains a minimum of $0$ at only the point $B$ (you show this by setting the things inside parentheses to equal 0). Since $f$ is an integer polynomial, for any integer point $X$ we have $f(X) \ge 1$ (since I showed it doesn't equal 0). Note that $f(A) = 1$. So it suffices to show that $f$ attains the value $1$ only at $A$. We firstly need for $(4x_1-1)^2 \le 1 \implies x_1 = 0$ (remember that everything is an integer point). So $(4x_1-1)^2 = 1$, so we need to $(8039x_1-x_2)^2 = 0 \implies x_2 = 0$. So the only integer point that achieves the value $1$ is $A$s, so we're done.

How about $f(x,y)=\left( (x^2+y^2-(2009.9^2+0.1^2))(x^2+y^2-\epsilon) \right)^2 $ where $\epsilon$ is insanely small. Of course, $A=0, B=(2009.9,0.1)$. Let $r=0.1^2+2009.9^2$, not an integer. First, notice that $f(2009.9,0.1)=0$ and so it's the smallest possible value $f$ attains. Now, we want $f(0,0)=\left( (\epsilon)r \right) ^2 < (X-\epsilon)^2(X-r)^2$ for any $X$ positive integer. So we want $|\frac{\epsilon}{X-\epsilon}| < |\frac{X-r}{r}|$. Notice that the the RHS is bounded below, while the LHS is bounded above, so we can make the LHS as small as we want with a correct $\epsilon$.

WLOG $A$ is the origin. We want a polynomial $f\in\mathbb{Z}[x_1,x_2]$ which attains its minimum over $\mathbb{Z}^2$ at $A$, such that its value at $B$ is smaller. The conditions are very weak, so yes, this is a construction problem. Say we just want a minimum at $A$. Then we can just take the distance from $A$ squared, $x^2+y^2$, that's minimal at $A$. The joke is that $B$ is not necessarily an integer point. For example, the polynomial $t^2-t$ is negative in $]0;1[$, but non-negative elsewhere. Seriously abusing this, we choose $B=(2009.5,0.5)$ and set \[f(x,y)=(x^2+y^2)+\text{googolplex}\cdot (y^2-y).\] Then $f(x,y)=0$ for $(x,y)=A$, but $f(x,y)>0$ for any other integer point. However, $f(B)$ is absurdly negative because $0.5^2-0.5=-\frac14$.

Define the polynomial $g(X,Y)$ as follows: $$g(X,Y)=(X^2+Y^2)\cdot \left(2\cdot (X^2+Y^2-2010^2+1)^2-1\right)$$and let us set $f(X,Y)=g(X-a_1,Y-a_2)$. Let $c,d$ be positive real numbers such that $c+d=2010$ and $c^2+d^2=2010^2-1$. It is clear that such numbers $c,d$ exist and neither is an integer. Let $b_1-a_1=c$ and $b_2-a_2=d$. Clearly, $g$ has integer coefficients and for all non zero lattice points $(a,b)$ we have $a^2+b^2>0$ and since $2010^2-1=2009\cdot 2011$ and $2011$ is a prime number of the form $4k+3$ with exponent $1$ in this expression, we see that $a^2+b^2 \not= 2010^2-1$. Thus, $g(a,b)>g(0,0)=0$ and as $c^2+d^2=2010^2-1$ we get that $g(c,d)<0$ and thus, $f$ satisfies all the desired conditions.

Let $g(t)=t(3t-(3k+1))(3t-(3k+2))$, where $k$ is a positive integer. Note that $g$ is minimized on nonnegative integer inputs at $t=0$ and minimized on nonnegative reals at some $t=a\in(3k+1,3k+2)$. In particular $g(t)>0$ for all $t\in\mathbb N$. Let $f(x,y)=g(x^2+y^2)$, and set $k$ to be some positive integer such that the circle $x^2+y^2=a^2$ intersects the diamond $|x|+|y|=2010$ at some point $(b_1,b_2)$. Then, we take $(a_1,a_2)=(0,0)$, and note that all conditions are satisfied.

Does this work? $f(x,y)=((2x-2011)^2+(2y-2009)^2)(C(x^2+y^2)+1)$ $A=(0,0)$ $B=(2011/2,2009/2)$ for $C$ sufficiently large