I know that you are thinking firmly about the integrable world, but I thought it worth adding that for symplectic manifolds, there is no obvious generalisation of Gromov-Witten theory to higher dimensional subvarieties. This is because to define "holomorphic" you use a non-integrable almost complex structure and non-integrability means that there are no higher dimensional holomorphic objects. The fact that there are holomorphic curves can be thought of as an instance of the fact that all almost complex structures over 2-manifolds are automatically integrable. (E.g., since there are no (2,0)-forms, the space where the Nijenhuis tensor should live is zero.)
It turns out that condition (T) is, indeed, sufficient for the $27$ lines (distinct and intersecting as expected) to lie on a cubic surface.
To see this, consider the lines $a_1,a_2,a_3,a_4,a_5$ and $b_6$, where the labeling is as in note (2) of the question: $a_1$ through $a_5$ are pairwise skew, and $b_6$ intersects all of them. Choose $4$ distinct points on $b_6$, and $3$ distinct points on each of $a_1$ through $a_5$ also distinct from the intersection point with $b_6$: this makes $4+5\times3=19$ points in total; since there are $20$ coefficients in a cubic form on $4$ variables, there exists a cubic surface $S$ passing through these $19$ points. Since $S$ contains four distinct points on $b_6$, it contains $b_6$ entirely, and since it contains four points (viz., the intersection with $b_6$ and the three additional chosen points) on each of $a_1$ through $a_5$, it contains these also.
Now $b_1$ intersects the four lines $a_2$ through $a_5$ in four distinct points since $a_2$ through $a_5$ are pairwise skew; and since these four points lie on $S$, it follows that $b_1$ lies entirely on $S$; similarly, $b_2$ through $b_5$, and finally $a_6$ (which intersects $b_1$ through $b_5$ in distinct points), all lie on $S$. So $S$ contains the "double-six" $\{a_1,\ldots,a_6,b_1,\ldots,b_6\}$.
So far, property (T) has not been used, only the incidence relation. Now it remains to see that the $c_{ij}$ lie on $S$. Consider the intersection line $\ell$ of the planes $a_i\vee b_j$ (spanned by $a_i$ and $b_j$) and $a_j\vee b_i$ (spanned by $a_j$ and $b_i$; these planes are well-defined since $a_i$ meets $b_j$ and $a_j$ meets $b_i$, and they are distinct since $a_i$ and $a_j$ are skew): this line $\ell$ must be equal to the $c_{ij}$ of the given configuration, because $c_{ij}$ intersects both $a_i$ and $b_j$ so property (T) implies that it lies in the plane $a_i\vee b_j$, and similarly it lies in the plane $a_j\vee b_i$. But $\ell$ also lies on $S$ for similar reasons¹, in other words, $c_{ij}$ lies on $S$, and all the given lines lie on $S$.
Finally, $S$ must be smooth because it contains the configuration of $27$ lines expected of a smooth cubic surface (it is easy to rule out the case where $S$ is a cubic cone, a reducible surface or a scroll by considering the intersecting and skew lines in the double-six; and every configuration where $S$ has double point singularities has fewer than $27$ lines).
- Let me be very precise here, because at this stage we don't know whether $S$ is smooth (so we can't invoke (T) directly on $S$). We have four distinct lines $a_i,a_j,b_i,b_j$ on $S$ such that $a_i$ meets $b_j$ and $a_j$ meets $b_i$ and all other pairs are skew. Call $\pi := a_i\vee b_j$ and $\pi' := a_j \vee b_i$ the planes generated by the two pairs of concurrent lines, and $\ell := \pi\wedge\pi'$ their intersection. We want to show that $\ell$ lies on the surface $S$. If $\pi$ or $\pi'$ is contained in $S$ (reducible) then the conclusion is trivial, so we can assume this is not the case. So the (schematic) intersection of $S$ with the plane $\pi$ is a cubic curve containing two distinct lines ($a_i$ and $b_j$), so it is the union of three lines: $a_i$, $b_j$ and a third line $m$ (a priori possibly equal to one of the former). Consider the intersection point of $a_j$ and $\pi$ (which is well-defined since $a_j$ is skew with $a_i$ so does not lie in $\pi$): it lies on $\ell$ because it is on both $\pi$ and $\pi'$; and it must also lie on $m$ since it is on $\pi$ but neither on $a_i$ nor on $b_j$ (as the two are skew with $a_j$); similarly, the intersection point $b_i\wedge\pi$ is well-defined and lies on both $\ell$ and $m$; so $\ell=m$ lies on $S$ (and we are finished) unless perhaps the two intersections considered are equal, i.e., $a_j,b_i,\pi$ concur at a point $P$, necessarily on $m$. Assume the latter case: $S$ must be singular at $P$ because the line $m$ through $P$ does not lie on the plane $\pi'$ generated by two lines ($a_j,b_i$) through $P$ (i.e., we have three non-coplanar tangent directions at $P$). Now symmetrically, if we call $m'$ the third line of the intersection of $S$ with $\pi'$ (besides $a_j$ and $b_i$), we are done unless $a_i,b_j,\pi'$ concur at a point $P'$, necessarily on $m'$ and necessarily singular on $S$. The points $P$ and $P'$ are distinct because $a_i$ and $a_j$ are skew; and they are on $\ell$ because they are on $\pi$ and $\pi'$; and the line joining two singular points on a cubic surface lies on the surface, so $\ell = P\vee P'$ lies on $S$ in any case. (Phew!)
What I still don't know is whether there are configurations of $27$ distinct lines with the expected incidence relations and which satisfy neither condition (T) nor its dual (viz., whenever three lines pairwise meet, all three meet at a common point), and in particular, what are the irreducible components of the space of configurations. (I also don't know if there is a way to substantially simplify the tedious argument given in note (1) above.)
Best Answer
Generalizing BCnrd's example, almost all surfaces S of degree d ≥ 4 in P^3 have Picard group = Z, hence every curve on a general S is a complete intersection with another surface. Thus any two curves on S intersect. This is the theorem of Noether-Lefschetz.
http://www.springerlink.com/content/t754510m417u0712/