Solution:
The train's distance from the man's original position is $t^{2} + d_{0}$, and the man's distance from his original position if he runs at speed $v$ is $v t$ at time $t$. We need to find where $t^{2} + d_{0} = v t$ has a solution. Note that this is a quadratic equation with discriminant $D = \sqrt{v^{2} - 4 d_{0}}$, so it has solutions for real $D$, i.e. where $v \geq \sqrt{4 d_{0}}$, so $4 d_{0}$ must be a perfect square. This happens when $4 d_{0}$ is an even power of $2$: the smallest value is $2^{0}$, the second smallest is $2^{2}$, the third smallest is $2^{4}$, and in general the $n$th smallest is $2^{2(n-1)}$, or $4^{n-1}$.