How can a twin travelling astronaut paradox be resolved? This is a variant of the twin paradox. But having each of the twin astronauts take off in opposite directions and returning to meet such that all aspects of acceleration and velocity are the same. Both should observe the other’s clock being slow for the whole trip but when they meet they should have aged equally. How can this be (or where is the error in this setup)?
 A: The mistake you are making is that you are forgetting about the relativity of simultaneity. Although each twin's clock will seem to tick slowly compared with clocks stationary in the other twin's reference frame, those other clocks will appear to be out of synch to the twin who is moving relative to them. Over the course of their identical return journeys, the effects of time dilation observed by the twins are exactly cancelled by the effects of the relativity of simultaneity, so the twins will have aged by the same amount when they meet again.
A: 
Both should observe the other’s clock being slow for the whole trip

This is the part you have got wrong.
In twin A's frame, twin B would age slower during the 1st half of twin A's journey
In twin A's frame, twin B would also age slower during the 2nd half (return half) of twin A's journey.
But that does not mean that twin B would age slower over the combined 1st + 2nd half.
This is because of what happens at the turnaround point.
In twin A's frame, twin B's age at the end of twin A's first half WOULD NOT be the same as twin B's age at the beginning of the 2nd half. There will be a discontinuity/time jump there.
Let's say, if you are one twin and i am another.
So the following would be how events would happen in your frame ( this is ONLY for your frame, in other frames it will be different )
In your frame, suppose your journey takes 10 + 10 = 20 years.
At the end of your 1st half you will be 10 years old, and i would be say, 5 years old, as i would indeed have aged slower in your frame.
Now when you turn around  and start your second half journey,, whereas you will be 10 years old, i would NOT be 5 years old. I would be 15 years old.
Then, during the 2nd half of your journey, you would age by 10 more years and i would age by 5 more years, as i would age slower as expected.
When we meet up you would have aged  by 10 +10 and i would have aged by 15 + 5 .
Hence, we would have the same age
A: This version just adds complexity to the normal Twin Paradox. In the normal version, the space twin ages $\frac 1{\gamma}\frac L{\beta}$ on each leg of the journey, but sees the Earth twin age only $\frac 1{\gamma^2}\frac L{\beta}$. Meanwhile,  during a leg, Earth sees earth age $\frac{L}{\beta}$.
The missing
$$2\times  \big(\frac L {\beta} -\frac 1{\gamma^2}\frac L{\beta}\big) = 2{L}{\beta}$$
(where the 2 is for both legs)
is exactly the clock bias change when the twin turns around. Since time transforms as:
$$ t'=\gamma(t-\beta x) $$
the time back on Earth (paradoxically) changes by:
$$t'(-\beta) -t'(+\beta) =[\gamma(t+\beta x)] - [\gamma(t-\beta x)]=2L\beta$$
Lorentz transforms are linear: there is a slope and an intercept. We look to the slope (the Lorentz factor) as the cause of paradoxes, when in fact it the intercept (velocity dependent bias of distant clocks) that causes them.
So now in your set up, you have 2 velocities:
$$\beta$$
and
$$ \beta'=\frac{2\beta}{1+\beta^2}$$
During the turn arounds, the other twin is moving  and changing reference frame...it's much more complicated.
But the resolution is the same: when you are far away ($L$) from a point, the time ($t_1'$) there depends on which way you are moving. If your clock says $t_0$, then all you can say is:
$$ (t_0 -\frac L c) \lt t_1'\lt( t_0 +\frac L c)$$
You can chose any time you want, if you can get to any velocity on $v\in(-c,+c)$ w.r.t to the direction of the point in question.
Edit: Nevertheless, since we know SR is correct, we can figure out what happens w/o doing the tedious part. On each leg, a twin ages $\frac{L}{\gamma \beta}$, and sees the other twin age $\frac{L}{\gamma'\gamma \beta}$ where
$$\gamma'=\frac 1{\sqrt{1-\beta'^2}}$$
which means the turn around jumps over:
$$\Delta t= 2\times [\frac L {\beta\gamma} -\frac L{\gamma'\gamma \beta}] $$
$$\Delta t=2\frac L{\beta\gamma}[1-\frac 1{\gamma'}]$$
