Finding the total angular momentum for a system of particles

It's just the third law ${bf F}_{ij}= -{bf F}_{ji}$, so $$ sum_{i,j} {bf r}_itimes {bf F}_{ij}= frac 12 sum_{i,j} ({bf r}_itimes {bf F}_{ij}+{bf r}_jtimes {bf F}_{ji}) quad (hbox{just relabelling }ileftrightarrow j) = frac 12 sum_{i,j} ({bf r}_itimes {bf F}_{ij}-{bf r}_jtimes {bf F}_{ij}) frac 12 sum_{i,j} ({bf r}_i-{bf r}_j)times {bf F}_{ij} = sum_{langle i,jrangle} ({bf r}_i-{bf r}_j)times {bf F}_{ij}, $$ where $langle i,jrangle$ denotes the pair composed of $i$ and $j$. This means that $langle 1,2rangle $ is not counted twice as being $i=1,j=2$ and $i=2,j=1$, for example.

I bet your book goes on to say that $({bf r}_i-{bf r}_j)times {bf F}_{ij}=0$ because the force is parallel to the line separating the particles. If the book does do this, it is a common textbook cheat because there is no reason for this to be so if the forces arise from chemical bonds. The sum $$ sum_{i,j}({bf r}_i-{bf r}_j)times {bf F}_{ij} $$ does vanish for a rigid body --- but the reason for this is not that the individual $({bf r}_i-{bf r}_j)$ are parallel to ${bf F}_{ij}$.

This "parallel" argument is correct, however, if the forces are gravitational or electrostatic, as these are examples of the special case of central forces.