Question: In Reid's commutative algebra, the author gives some exposition about the Nullstellensatz, $I(V(J))=\operatorname{rad}(J)$. But I can't understand it: The Nullstellensatz says that we can take the intersection just over maximal ideals of $k[X\_1,\cdots,X\_n]$.

Answer: 
In a commutative ring $A$, an ideal $I\subset A$ has two radicals:  

a) The nilpotent radical $\sqrt I=Rad(I)=\cap\_{I\subset P \:prime}$ which is the intersection of all prime ideals containing $I$.  

b) The Jacobson radical $J(I)=\cap\_{I\subset M \: maximal}$ which is the intersection of all maximal ideals containing $I$.  

Now, there is a class of rings, deservedly called *Jacobson* rings, for which the two radicals coincide. Any algebra of finite type over a field is a Jacobson ring, and so there is no contradiction in the statements you mention in your question.

*Warning* Many rings are not Jacobson rings. For example a local domain (A,M) which is not a field is never Jacobson because the nilpotent ideal $(0)$ has $Rad((0))=(0)$ whereas $J((0))=M$.  

So, beware: the local rings $\mathcal O\_{X,x}$ of points $x\in X$ of the affine algebraic $k$-variety $X=Spec(A)$ are *not* Jacobson rings even though the finitely generated $k$-algebra $A$ *is* a Jacobson ring!
Because the maximal ideals correspond to points in $k^n$. Then $f(a\_1,\ldots,a\_n)=0$ iff $f \in (X\_1-a\_1,\ldots,X\_n-a\_n)$.