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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(0)-x_3(0))^3}$, call this $A_4$, and similarly $\dfrac{dx_1}{dt} \ge - A_1 = -\dfrac{m_1}{(x_2(0) - x_1(0))^3}$. So $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_3 - x_2)}$$ $$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_2 - x_3)}$$ $$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} $$ Thus a collision will occur by time $T$ if $$ \int_0^T \dfrac{ dt}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$ The integral of the left side from $0$ to $\infty$ is finite, but doesn't depend on $m_2$ or $m_3$. So certainly if $m_2$ or $m_3$ is large enough (depending on $m_1$, $m_2$ and the $x_j(0)$) a collision will occur in finite time.