In general, this is not possible for all distances. If the lines do not cross, there willl be a minimum distance between the lines which you can never reach below, but you should be able to solve it for all distances equal to or larger than this minimum distance.
Since the general case has been shown by Eivind, I will take the special case where the lines do cross, so a solution is guaranteed.
If the lines, say $l$ and $m$, do cross, you can write them as:
l(t)&= \mathbf a + t \mathbf b \\
m(s)&= \mathbf a + s \mathbf c
The distance between two lines will be $\|l(t) - m(s)\| = \| t \mathbf b - s \mathbf c\|$. Say you want this to be equal to $d$, then you can solve for $d^2$ (I am here assuming that you are working in $\mathbb R^3$):
d^2 &= \langle \mathbf b t - \mathbf c s, \mathbf b t - \mathbf c s \rangle = \langle \mathbf b t, \mathbf b t - \mathbf c s \rangle - \langle \mathbf c s, \mathbf b t - \mathbf c s \rangle \\
&= \langle \mathbf b t, \mathbf b t \rangle - 2 \langle \mathbf b t, \mathbf c s \rangle + \langle \mathbf c s, \mathbf c s \rangle = \\
&= t^2\|b\|^2 + s^2 \|c\|^2 - 2 \langle \mathbf b t, \mathbf c s \rangle
If you have normalized $\mathbf b$ and $\mathbf c$, this will reduce to:
$$d^2 = t^2 + s^2 - 2 (b_1c_1 + b_2c_2 + b_3c_3) ts$$
Which you can solve. You can here choose a value for $t$ (which will tell you how far the crossing point the point on line $l$ is) and then solve for $s$. If you want the points to be on equal distance from the crossing point, take $t = s$.