If X is a smooth projective curve over an algebraically closed field k, then we can make a (huge, useless) abelian group Div(X) which is the set of formal sums of points on X. (The "free abelian group" on X).

It would be flippant to say Div(X) is an answer to your question, since it has nothing to do with geometry at all (we can form the free abelian group on any set). An element of Div(X) looks like \sum n_i P_i where n_i are integers and P_i are points on X, and they "add" in the obvious way. The sum doesn't "mean" anything. But we can get to geometry from it.

Inside Div(X) there is a subgroup, Div^0(X), of formal sums of points such that the set of coefficients is zero. Still nothing to do with geometry.

Inside Div^0(X), there is a very interesting subgroup, which is the set of "divisors of functions." Namely, if f is a rational function on X (meaning it's locally a quotient of polynomials), we get an element of Div^0(X) by taking \sum P_i - \sum Q_i where P_i are the zeroes of f and Q_i are the poles (caveat - you have to count them with multiplicity). This is an element of Div(X) but is not obviously an element of Div^0(X) -- this uses the fact that X is projective. Let's call the subgroup that comes this way Princ(X) (for "principal" divisors).

Now we get an interesting group that does have something to do with geometry, which is called Pic^0(X), by taking the quotient Div^0(X)/Princ(X).

Amazing theorem: there is a natural isomorphism from X to Pic^0(X) if and only if X is of genus one, i.e. an elliptic curve. (In general, Pic^0(X) is an abelian variety whose dimension is the genus of the corresponding curve.) This is why only elliptic curves (among the projective ones) are "naturally" groups. The relationship with the usual picture with the lines is that the intersection locus of the lines is the principal divisor associated with a functional that vanishes along the line.