= Solution
About a fixed inertial origin, the total <angular momentum> is $\boldsymbol L=\sum_i m_i\boldsymbol r_i\times\dot{\boldsymbol r}_i$. Differentiation gives
$$
\dot{\boldsymbol L}=\sum_i\boldsymbol r_i\times m_i\ddot{\boldsymbol r}_i,
$$
because $\dot{\boldsymbol r}_i\times\dot{\boldsymbol r}_i=0$. Separate the external and internal contributions. The combined <torque> of an internal pair is
$$
\boldsymbol r_i\times\boldsymbol F_{ij}
+\boldsymbol r_j\times\boldsymbol F_{ji}
=(\boldsymbol r_i-\boldsymbol r_j)\times\boldsymbol F_{ij}=0.
$$
The first equality uses <Newton's third law>, and the second uses the given central-force condition. All internal torques therefore cancel, leaving
$$
\boxed{\frac{d\boldsymbol L}{dt}=\boldsymbol N,
\qquad\boldsymbol N=\sum_i\boldsymbol r_i\times\boldsymbol F_i^e.}
$$
An equal-and-opposite pair which is not central need not have zero combined <torque>; this explains why both internal-force hypotheses matter.
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