Does every finitely generated group have a maximal normal subgroup? Given an infinite group which is finitely generated, is there a proper maximal normal subgroup?
 A: In general, it is true for quite general structures: finitely generated ones have a maximal proper substructure as soon as they have at least proper substructure.
Namely consider universal-algebraic structure $G$: this means a set $G$ endowed with a given family of finitary laws $(\mu_i)_{i\in I}$. A substructure is a subset which is stable under every law (which in particular means it contains all constants, i.e., all images of 0-ary laws).
Write $G_0$ for the substructure generated by the set of such constants (the set of elements that are image of some given 0-ary law $\mu_i$). For instance, for a group or monoid $G$, we have $G_0=\{1_G\}$; for a semigroup $G$ we have $G_0=\emptyset$; for a unital ring $R$, we have $R_0=\mathbf{Z}1_R$, etc. Then $G$ has at least a proper substructure iff $G\neq G_0$.
The structure $G$ is said to be generated by a subset $Y$ the only substructure containing $Y$ is $G$; it is finitely generated if it is generated by some finite subset.
Now if $G$ is finitely generated (by some subset $Y$) and $G\neq G_0$, then Zorn applies to the set of proper substructures (indeed for a substructure $H$, the condition $H\neq G$ means that $H\cap Y\neq Y$, and this condition passes to increasing unions). Hence:

If $G$ is a finitely generated structure (with respect to this family of laws) and $G\neq G_0$ then $G$ has a maximal proper substructure.

This applies in particular to groups. For a group $G$, we have 3 laws (binary product, unary inverse, zero-ary unit), so $G_0=\{1_G\}$.

Hence every finitely generated group $G\neq\{1\}$ has a maximal proper subgroup.

But given a fixed group $G$, this also applies to $G$-groups: these are groups $H$ endowed with an action of $G$ (thus $I$ is formed of the 3 laws of the group $H$, and one unary law for each $g\in G$ defining the action of $G$, plus axioms that are not important here, saying that this is a $G$-action. This applies to $G$ endowed to the $G$-action by conjugation:

If $G\neq\{1\}$ is a group and $G$ is finitely generated as normal subgroup of itself (e.g., $G$ is finitely generated as group), then $G$ has a maximal proper normal subgroup.

[Remark: for an abelian group $A$, there is a maximal proper subgroup in $A$ iff $A$ is not divisible, i.e., iff there exists $n\ge 2$ such that $A\neq nA$.]
A: So many answers! I'm completely lost. The paper of "B.H. Neumann, "Some remarks on infinite groups", Journal London Math. Soc, 12 (1937), 120-127" stated results for the existence of maximal subgroups, not maximal normal subgroup. Is this existence question of nontrivial normal subgroup still unsolved?
A: If you mean nontrivial maximal normal subgroup (not 1 or the whole group), then the answer is no.
Higman constructed a finitely generated infinite group $G$ with no subgroups of finite index.  You then get a finitely generated group with no nontrivial normal subgroups by taking the quotient by a maximal normal subgroup.
Higman's group $G$ is $\langle a,b,c,d | a^{-1} b a = b^2, b^{-1}cb = c^2, c^{-1}dc=d^2, d^{-1}ad=a^2 \rangle$
See Higman, Graham. A finitely generated infinite simple group. J. London Math. Soc. 26, (1951). 61--64. 
Edit:
If you mean does it have a proper maximal normal subgroup, then the answer is yes:
Finitely generated groups have a (possibly trivial) maximal normal subgroup. Higman's reference for this is B.H. Neumann, "Some remarks on infinite groups ", Journal London Math. Soc, 12 (1937), 120-127.
A: Assuming you mean "does a maximal normal subgroup always exist?" (and that you don't care about computing it), here is a way to restate the problem. Notice that if G has no maximal normal subgroups, that means that every proper normal subgroup H of G is contained in a larger proper normal subgroup K of G. In particular, this means that the group G/H must not be finite; if it were, we could only find a finite chain of normal subgroups between H and G. So the question "does a maximal normal subgroup always exist" is the same as "must a finitely generated group have any finite nontrivial quotients?" I'm not sure what the answer to that is, but it seems like a useful restatement.
A: Check out the Tarski monster. It is 2-generated and simple.
Unless I misunderstood your question and you exclude infinite simple groups altogether.
