I am also unsure of what "nontriviality" conditions you want to impose. Without any further conditions, the following answers your question:
Call a positive integer $n$ nilpotent if every group of order $n$ is nilpotent.
Call a positive integer $n$ abelian if every group of order $n$ is abelian.
Suppose that the prime factorization of $n$ is $p_1^{a_1} \cdots p_r^{a_r}$. Then:
$n$ is nilpotent iff for all $i,j,k$ with $1 \leq k \leq a_i$, $p_i^k \not \equiv 1 \pmod{p_j}$.
$n$ is abelian iff it is nilpotent and $a_i \leq 2$ for all $i$.
These results are proved in
Pakianathan, Jonathan(1-WI); Shankar, Krishnan(1-MI)
Nilpotent numbers.
Amer. Math. Monthly 107 (2000), no. 7, 631--634.
The proofs are constructive: for any $n$ which is not nilpotent (resp. abelian), they give an explicit group of that order which is not nilpotent (resp. abelian).
The paper is available at
http://alpha.math.uga.edu/~pete/nilpotentnumbers.pdf
I reject the premise of the question. :-)
It is true, as Terry suggests, that there is a nice dynamical proof of the classification of finite abelian groups. If $A$ is finite, then for every prime $p$ has a stable kernel $A_p$ and a stable image $A_p^\perp$ in $A$, by definition the limits of the kernel and image of $p^n$ as $n \to \infty$. You can show that this yields a direct sum decomposition of $A$, and you can use linear algebra to classify the dynamics of the action of $p$ on $A_p$. A similar argument appears in Matthew Emerton's proof. As Terry says, this proof is nice because it works for finitely generated torsion modules over any PID. In particular, it establishes Jordan canonical form for finite-dimensional modules over $k[x]$, where $k$ is an algebraically closed field. My objection is that finite abelian groups look easier than finitely generated abelian groups in this question.
The slickest proof of the classification that I know is one that assimilates the ideas of Smith normal form. Ben's question is not entirely fair to Smith normal form, because you do not need finitely many relations. That is, Smith normal form exists for matrices with finitely many columns, not just for finite matrices. This is one of the tricks in the proof that I give next.
Theorem. If $A$ is an abelian group with $n$ generators, then it is a direct sum of at most $n$ cyclic groups.
Proof. By induction on $n$. If $A$ has a presentation with $n$ generators and no relations, then $A$ is free and we are done. Otherwise, define the height of any $n$-generator presentation of $A$ to be the least norm $|x|$ of any non-zero coefficient $x$ that appears in some relation. Choose a presentation with least height, and let $a \in A$ be the generator such that $R = xa + \ldots = 0$ is the pivotal relation. (Pun intended. :-) )
The coefficient $y$ of $a$ in any other relation must be a multiple of $x$, because otherwise if we set $y = qx+r$, we can make a relation with coefficient $r$. By the same argument, we can assume that $a$ does not appear in any other relation.
The coefficient $z$ of another generator $b$ in the relation $R$ must also be a multiple of $x$, because otherwise if we set $z = qx+r$ and replace $a$ with $a' = a+qb$, the coefficient $r$ would appear in $R$. By the same argument, we can assume that the relation $R$ consists only of the equation $xa = 0$, and without ruining the previous property that $a$ does not appear in other relations. Thus $A \cong \mathbb{Z}/x \oplus A'$, and $A'$ has $n-1$ generators. □
Compare the complexity of this argument to the other arguments supplied so far.
Minimizing the norm $|x|$ is a powerful step. With just a little more work, you can show that $x$ divides every coefficient in the presentation, and not just every coefficient in the same row and column. Thus, each modulus $x_k$ that you produce divides the next modulus $x_{k+1}$.
Another way to describe the argument is that Smith normal form is a matrix version of the Euclidean algorithm. If you're happy with the usual Euclidean algorithm, then you should be happy with its matrix form; it's only a bit more complicated.
The proof immediately works for any Euclidean domain; in particular, it also implies the Jordan canonical form theorem. And it only needs minor changes to apply to general PIDs.
Best Answer
See:
V. S. Guba, Finitely generated complete groups, Izv. Akad. Nauk SSSR Ser. Mat. 50 (1986), 883-924.
for an interesting 2-generated example. (Furthermore, roots in Guba's examples are unique.) Note that a group $G$ is called complete if for any non-trivial word $u(x_1,\cdots,x_m)$ and every $g\in G$, the equation $u(x_1,\cdots,x_m)=g$ has a solution. In your case, the words $u$ are of the form $x^n$. I do not know if Guba's group is simple, but it does have a (nontrivial) simple quotient, which is necessarily verbal, and, in particular, divisible.
None of these groups is finitely-presented and, I think, the problem of existence of fp divisible groups is open. The philosophical reason why is the following. Call an infinite finitely-generated group "exotic" if it satisfies some bizarre, seemingly impossible property, e.g., being a torsion group, containing very few conjugacy classes, being divisible, etc. The most common method for constructing exotic groups $G$ is by direct limit of a sequence of (relatively) hyperbolic groups $G_k$ which are quotients of a single group $G_0$. If $G$ were finitely presented, it would be isomorphic to one of the groups $G_k$ and, hence, non-exotic.
Mark Sapir will probably have more comments on this.