Resistance of a fully conected graph?
The equal-resistance case

In the case where all resistances are equal (call it \$R\$), the solution is straightforward. Without loss of generality we can label the two nodes we are measuring between as node 1 and node 2.  We consider a voltage applied between the two nodes. We can re-draw the graph with the entry and exit nodes separate, and the others in a line:







simulate this circuit &ndash; Schematic created using CircuitLab

Here I have omitted many resistors which joins nodes 3 to n with each other.  However, by symmetry, it should be obvious that nodes 3 to n are all at the same potential, so no current flows between them and the resistors joining them are irrelevant.  It should now be obvious that the resistance between Node 0 and Node 1 is the parallel combination of R and N-2 instances of two resistors in series.  Thus the total resistance is:

\$R_{1,2} = \left(\frac{1}{R} + \frac{(n-2)}{2R}\right)^{-1} = \frac{2R}{n}\$

The arbitrary resistance case

There is (as far as I know) no simple solution.  You must assume some voltage between the nodes of interest, and then write down ohms law for each resistor in the graph, and current conservation equations for each node.  For a graph with \$n\$ nodes, you will get \$n\$ current conservation equations, and \$n^2\$ resistance equations. You now have \$n^2+n\$ equations, and the \$n^2+n-1\$ unknowns, so you can solve the system of linear equations. This may be quite time consuming.