Here is my own rule of thumb about this.

One important example of a monadic category over $C$ is one obtained by adjoining to $C$ a system of $n$-ary operations satisfying some relations. For example, the category of rings with involution is monadic over sets: it can be given by two binary operations ($+,\times$), two $0$-ary operations ($0,1$), and two $1$-ary operations (negation and the involution). Perhaps if you adopt a suitably enlightened definition of operations with relations then all monadic functors would arise in this way.

Now a $1$-ary operation is just a morphism, so you can also view it as a $1$-ary operation in the opposite category. So if you're adding only $1$-ary operations, then you should get a category which is both monadic and comonadic over the original category.

Indeed, the usual way of defining lambda-rings is by adjoining a bunch of $1$-ary operations to the category of commutative rings. The monad is the free lambda-ring functor, and the comonad is the big Witt vector functor (also called the co-free lambda-ring functor). Another example is the category formed from objects of the original category $C$ together with an action of your favorite group (or monoid) $G$. In representation theory, the monad is often called the induced representation functor, and the comonad is called the co-induced representation functor.

I'd expect that, as above, with suitably enlightened definitions, all examples would arise in this way.

Some things that follow formally from this set up are that the new category has all the same kinds of limits and colimits that the original category $C$ has, and the forgetful functor preserves them. Beck's theorem says that some kind of converse is also true.

morallye.g what's a nice mantra about bimonadic functors. $\endgroup$ – Arrow Jun 26 '17 at 14:48