= Solution
For distinct positions and positive masses, <Newton's law of universal gravitation> gives
$$
m_i\ddot{\mathbf x}_i=-G\sum_{j\ne i}m_im_j\frac{\mathbf x_i-\mathbf x_j}{|\mathbf x_i-\mathbf x_j|^3}.
$$
The proposed <homothetic gravitational motion> has $\ddot{\mathbf x}_i=-(2/9)t^{-4/3}\mathbf a_i$ and pair distances $t^{2/3}|\mathbf a_i-\mathbf a_j|$, so cancellation of $t^{-4/3}$ gives the time-independent equations
$$
\boxed{\frac29\mathbf a_i=G\sum_{j\ne i}m_j\frac{\mathbf a_i-\mathbf a_j}{|\mathbf a_i-\mathbf a_j|^3}}.
$$
These define a <Newtonian central configuration> with this choice of scale. For
$$
\Phi=\frac19\sum_i m_i|\mathbf a_i|^2+\sum_{i<j}\frac{Gm_im_j}{|\mathbf a_i-\mathbf a_j|},
$$
<differentiation> in $\mathbf a_i$ gives $\nabla_{\mathbf a_i}\Phi=(2/9)m_i\mathbf a_i-G\sum_{j\ne i}m_im_j(\mathbf a_i-\mathbf a_j)/|\mathbf a_i-\mathbf a_j|^3=0$, exactly the displayed equations. The double sum in the paper counts each pair twice, so it agrees with this expression.
Multiply the configuration equations by $m_i$ and sum. Pair contributions cancel, giving $\sum_i m_i\mathbf a_i=0$. Thus the total <linear momentum> is $\mathbf P=(2/3)t^{-1/3}\sum_i m_i\mathbf a_i=0$. Each position is parallel to its <velocity>, so every $m_i\mathbf x_i\times\dot{\mathbf x}_i=0$ and the total <angular momentum> is zero. About any other fixed point $\mathbf b$, it is $\mathbf L_{\mathbf b}=\mathbf L_O-\mathbf b\times\mathbf P=\boxed{\mathbf0}$.
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