Assume $x_1(0) < x_2(0) < x_3(0) < x_4(0)$.  Note that $\dfrac{dx_2}{dt}$ and $\dfrac{dx_4}{dt}$ are positive and the other two are negative.  So for $t > 0$ (and before the collision), $x_1 < x_1(0) < x_2(0) < x_2 < x_3 < x_3(0) < x_4(0) < x_4$.  Now
$\dfrac{dx_4}{dt} \le \dfrac{m_4}{(x_4-x_3(0))^3}$.  Solving the differential equation obtained by making this an equality, we find that
 $x_4(t) \le x_3(0) + ((x_4(0) - x_3(0))^4 + 4 m_4 t)^{1/4}$.  Call the right side $B_4(t)$.
Similarly $x_1(t) \ge B_1(t) = x_2(0) - ((x_2(0) - x_1(0))^4 + 4 m_1 t)^{1/4}$. 


So $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_3 - x_2)}$$
$$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_2 - x_3)}$$
$$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))} $$
Thus a collision will occur by time $T$ if
$$ \int_0^T \dfrac{ dt}{(x_3(0) -B_1(t))(B_4(t) - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$
The integral of the left side from $0$ to $\infty$ is infinite, since $x_3(0) - B_1(t)$ and $B_4(t) - x_2(0)$ only grow like $t^{1/4}$ as $t \to \infty$.

So there will always be a collision in finite time.