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 $x^n$.

Mark Sapir will probably have more comments on this.

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.

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 from what I understand.) 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 $x^n$.

Mark Sapir will probably have more comments on this.

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 $x^n$.

Mark Sapir will probably have more comments on this.