Why group needs zero and negative number The way a Group is defined, it needs to contain zero and negative elements. 
Why do we need these restrictions, e.g. natural numbers would not satisfy this criterion.
 A: The Simple Answer
It's a definition, so it can be whatever we want.
The Historical Answer
The definition of a group was chosen for specific reasons. Although I haven't studied the history of group theory in enough detail to be an authority, I do have some idea how the abstract definition of a group evolved. The first groups to be studied were permutation groups. A significant body of theory had been proven for permutation groups long before the abstract definition of a group had ever been thought of, including:


*

*Lagrange's theorem, arguably the most basic theorem in modern group theory. Lagrange first proved an extremely special case of this (the action of $S_n$ on a rational function of $n$ variables) in a paper about solving fifth degree equations (see here for more information). It was slowly generalized over the course of the nineteenth century, this paper is a good overview.

*The definition of the order of an element, the modern definition of conjugation, and the  characterization of conjugacy classes in a permutation group are due to Cauchy (he wrote an overview of permutation thory in the 1850s or 40s).

*Cauchy's theorem that a group whose order is divisible by a prime $p$ must have an element of order $p$.

*Sylow proved his famous theorems for the special case of permutation groups.

*I believe the basis theorem for finite abelian groups was formulated for permutation groups before being proven with an abstract definition, although the documents I have are a little less clear on this point.


When abstract definitions of groups started cropping up in the late 19th century, it was because structures other than permutation groups had been showing up that exhibited similar properties. Since those structures all have inverses, inverses are included in the list of requirements. As you point out, there are many natural algebraic structures that don't have inverses, but those weren't the ones that interested mathematicians at the time. Why not? Well, because those structures (such as $\mathbb N$) don't exhibit the kinds of mathematical phenomena listed above, which leads us to...
The Mathematical Answer
The inclusion of inverses is necessary so as to make the above theorems true. Abstract definitions of groups started cropping up in the late 19th century, after all of the above (and probably more) had been established for permutation groups (and possibly other kinds of groups). These new axiomatic definitions were surely chosen so as to make the already existing body of theory work.
Here's a case study example: Lagrange's theorem. Let's step through the proof and see where we use the existence of inverses.
Let $G$ be a structure satisfying all the properties of a group, except possibly the existence of inverses (so, a monoid). Let $H$ be a sub-structure of $G$. We can consider the cosets of $H$, that is, $xH$ where $x\in G$. Lagrange's theorem states that (1) they are all the same size as $H$, and (2) they don't intersect.
For (1), let $xH$ be a coset. Every element in that coset is of the form $xh$, with $h\in H$, so we can associate each element in $xH$ to a different element of $H$ (they're all different because if $h=h'$ then $xh=xh'$). So $H$ is at least as big as $xH$ (so far we haven't used inverses). The problem is that we haven't come up with a way to assign a unique element of $H$ to each element of $xH$. There could be several $h\in H$ giving the same element $xh$. We need the map $h\to xh$ to be injective. This is implied by the existence of inverses, although I admit it doesn't necessarily require them. Wait for part (2).
For part (2), let $xH$ and $yH$ be two cosets. Suppose they intersect, so $a\in xH\cap yH$. Then there are $h,h'\in H$ such that $a=xh=yh'$. To conclude that $xH$ and $yH$ are the same coset, we would normally multiply on the left by $h^{-1}$ and get $x=y(h'h^{-1})$, so that $x\in yH$, $x=yh$ (for some $h\in H$), and $xH=y(hH)$. We can then conclude that $hH=H$, which requires that $H$ be closed and contain an identity but not necessarily inverses.
While this isn't a proof that any monoid satisfying Lagrange's theorem is a group (which might be true, but I'm not sure), I thought that simply showing how the typical proof of Lagrange's theorem uses inverses would be more pedagogically valuable.
A: To specify a group structure, you need to have a set $S$ and a binary operation $*$ on elements of $S$ that satisfy some basic axioms.  It is misleading to say "a group...needs to contain zero and negative elements."  That may be the case if the binary operation is addition, which it need not be:  for instance, multiplicative groups have a "1" and "reciprocal elements" rather than a "0" and "negative elements," and still other choices of $(S,*)$ can result in rather interesting embodiments of such concepts.
What a group must have is an identity element and the existence of an inverse for each member of the group.  These are two of the four group axioms; the other two are associativity and closure under the group operation.
Structures that satisfy a proper subset of these axioms are called various names depending on which axioms are (and are not) satisfied.  But they are not called groups.  So your question, "why does a group have to have an identity and inverses," really has a simple answer:  because that is what we agree to call objects that satisfy these properties.
A: Your question can be thougt as why do people need to define group in that way ? 
I think the basic reason is the simple equation $$ax=b$$
You want to create a system that this equation has always a uniqe solution in a set $G$.


*

*to have solution $ax=a$, we require the identity,

*to get $x=a^{-1}b$, we require $a^{-1}$ exist and uniqe..


Now, we are ready to create  algebraic system as our algebraic equation $ax=b$ make sense.
