System of $N$ particles that interact via a gravitational force I'm having some trouble starting this homework problem, I'm just not sure where to start. Any hints would be appreciated.

Consider a system of $N$ particles that interact via an attractive 'gravitational' force that is proportional to the distance between particles:
  $$ \mathbf F_{ji} = -km_im_j(\mathbf r_i - \mathbf r_j) $$
  where k is a postive constant and $i,j=1,2,...,N$. Determine the trajectories of the particles.  (Hint: Choose a reference frame in which the CM is at rest at the origin.)

Thank you.
 A: I am not sure this is a typo, but the force $\mathbf{F}_{ji}$ is definitely not gravitational, it should be of the form $F_{ij} \sim 1 / |\mathbf{r}_i - \mathbf{r}_j|$. Anyway, define the position of the center of mass as 
$$
\mathbf{R} = \frac{\sum_k m_k \mathbf{r}_k}{\sum_k m_k} = \frac{1}{M}\sum_k m_k \mathbf{r}_k \tag{1}
$$
Since the reference system is located at this location, we can set $\mathbf{R} = 0$. Besides this, note that
$$
\mathbf{F}_{ii} = 0 \tag{2}
$$
therefore, the force on the $i$-th particle is simply
\begin{eqnarray}
\mathbf{F}_i &=& \sum_{j\not= i} \mathbf{F}_{ji} \stackrel{(2)}{=} \sum_j \mathbf{F}_{ji} \\
&=& - k m_i \sum_j m_j (\mathbf{r}_i - \mathbf{r}_j) = -km_i \left(\sum_j m_j \mathbf{r}_i - \sum_j m_j \mathbf{r}_j\right) \\
&\stackrel{(1)}{=}& -km_i \left(M\mathbf{r}_i - M\mathbf{R} \right) = -km_iM \mathbf{r}_i \tag{3}
\end{eqnarray}
Newton's second law is then $\mathbf{F}_i = m_i\ddot{\mathbf{r}}_i$
$$
\ddot{\mathbf{r}}_i = -k M \mathbf{r}_i
$$
which can be easily solved
