= Solution
For the first particle, the derivative of its <angular momentum> about the origin is its <torque>:
$$
\frac d{dt}\left(m_1\mathbf q_1\times\dot{\mathbf q}_1\right)
=\mathbf q_1\times\left(-\frac{f'(r)}r(\mathbf q_1-\mathbf q_2)\right)
=\frac{f'(r)}r\mathbf q_1\times\mathbf q_2.
$$
This need not vanish. For example take $f(r)=r^2/2$ and instantaneous positions $\mathbf q_1=(1,0,0)$, $\mathbf q_2=(0,1,0)$; the <torque> is $(0,0,1)$. Therefore \b[it is not conserved in general], although it is in special configurations or for a constant potential.
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