= Solution
Work in an <inertial frame>, take all masses to be constant, and assume the internal forces obey the pairwise relation from <Newton's third law> $F_{ij}=-F_{ji}$. For the <angular momentum> result, also assume these pair forces are central, so $F_{ij}$ is parallel to $r_i-r_j$ and its pair torque vanishes. The equations from <Newton's second law> are
$$
m_i\ddot r_i=F_i+\sum_{j\ne i}F_{ij}.
$$
Summing them and pairing the internal forces gives
$$
\boxed{\frac{dP}{dt}=F,\qquad
P=\sum_i m_i\dot r_i,\qquad F=\sum_iF_i.}
$$
About a fixed point $a$, define the total <angular momentum> and external <torque> by
$$
L_a=\sum_i(r_i-a)\times m_i\dot r_i,\qquad
G_a=\sum_i(r_i-a)\times F_i.
$$
Differentiation produces no velocity term because $\dot r_i\times m_i\dot r_i=0$. Grouping internal terms into pairs gives
$$
\frac{dL_a}{dt}
=G_a+\sum_{i<j}\bigl[(r_i-a)\times F_{ij}+(r_j-a)\times F_{ji}\bigr]
=G_a+\sum_{i<j}(r_i-r_j)\times F_{ij}.
$$
Each final <cross product> is zero by centrality, proving $\boxed{dL_a/dt=G_a}$. Equal and opposite internal forces give <momentum conservation> when the total external force is zero, but do not in general make their internal <torques> cancel; the central-force assumption supplies that extra step.
Let $M=\sum_i m_i$ and let the <centre of mass> be $R=M^{-1}\sum_i m_i r_i$, so $P=M\dot R$. For an arbitrary moving reference point $a(t)$, the same differentiation gives
$$
\frac{dL_{a(t)}}{dt}=G_{a(t)}-\dot a\times P.
$$
Taking $a=R$, the extra term is $-\dot R\times M\dot R=0$. Therefore the corresponding <angular momentum about the centre of mass> satisfies
$$
\boxed{\frac{dL_R}{dt}=G_R.}
$$
Its definition using velocities relative to $\dot R$ agrees with the one above, since $\sum_i m_i(r_i-R)=0$ makes $\sum_i(r_i-R)\times m_i\dot R=0$. Thus this result remains true even when the <centre of mass> accelerates.
Finally take the common mass to be $m$ and $F_i=-k\dot r_i$, with constant $k$. About either a fixed point or the <centre of mass>,
$$
G=-k\sum_i(r_i-a)\times\dot r_i=-\frac km L.
$$
The angular-momentum equation becomes $\dot L=-(k/m)L$. Multiplying by the integrating factor $e^{kt/m}$ gives $\frac{d}{dt}(e^{kt/m}L)=0$, hence the <angular momentum decay under uniform linear drag> is
$$
\boxed{L(t)=L(0)e^{-kt/m}.}
$$
The common mass and common drag coefficient ensure that the same decay factor applies to every term.
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