No spatial ordering in 1+1D flat Minkowski spacetime? Consider 3 point masses in 1+1D flat Minkowski spacetime initially at rest w.r.t. each other and equally separated:
$A----B----C$
Suddenly, $A$ acquires a speed towards $C$ of $0.97c$. This corresponds to a Lorentz factor $\gamma(v) \simeq 4$.
From $A$'s viewpoint, distances to $B$ and $C$ are contracted by $4$:
$A-B-C$
No problem so far. From $B$'s viewpoint, distance to $C$ remains the same, but distance to $A$ is contracted by the same factor:
$A-B----C$
Still fine, just length contraction. Now comes the hard part. From $C$'s viewpoint, distance to $B$ is unaltered, but distance to $A$ is reduced by $4$:
$B--A--C$
So now $A$ lies in between $B$ and $C$.
Is this correct? Can observers disagree on which object is further away?
 A: Let's start with some basic


*

*A pair of distinct events $P$ and $Q$ can be space-like separated $(\Delta \vec{x})^2 > (c \Delta t)^2$, time-like separated $(\Delta \vec{x})^2 < (c \Delta t)^2$, or light-like separated $(\Delta \vec{x})^2 = (c \Delta t)^2$ (and there are many variant names for this conditions including "null separated"). And these categories are invariant: they are the same to all inertial observers.

*Events that are space-like separated have no guaranteed temporal ordering, but do have a guaranteed spacial ordering. They also do not have zero spacial separation in any frame, but  zero temporal separation in some frame.

*Events that are time-like separated have a guaranteed temporal ordering but no guaranteed spacial ordering. They also have zero spacial separation in some frame, but do not have zero temporal separation in any frame.

*Events that are light-like separated have both orderings guaranteed, and have non-zero spacial and temporal separation in all frame.
Now let's apply this understanding to your situation. You have assumed that there exist a single time (in their common reference frame) when the objects have a particular spacial ordering. 
Let's call the events marking the location of the objects at that moment (in the selected frame) $A_1$, $B_1$ and $C_1$. Because these happens at a common time in some frame these are necessarily space-like separated. So they have a fixed spacial ordering.
And the same can be said of any group of events that are simultaneous in the selected frame or in the frame of object $A$ after it has accelerated.
To find some events that might not have a unique spacial ordering consider event $A_2$ which occurs a non-trivial time after object $A$ accelerates. If we wait long enough then it will be space like separated from at least event $B_1$ and perhaps event $C_1$ as well.
A: No, only the first Lorentz contraction is correct. When A accelerates suddenly to $.97~c$, its position doesn't suddenly change in the frame of B and C. It's still 4 units to the left of B and 8 to the left of C.
Lorentz transformations relate measurements as made in different frames. B and C have not changed frames, so it is not appropriate for them to use a Lorentz transformation. A has changed frames, so a Lorentz transformation is appropriate. 
You don't use a Lorentz transformation whenever an object accelerates- you use a Lorentz transformation when an observer accelerates, or to relate the measurements made by two observers in different frames.
Two observers can, in general, disagree on spatial ordering in certain cases. But B and C are in the same frame, and so share a coordinate system. In this case, they certainly cannot disagree on spatial ordering.
A: A spacetime diagram might be helpful here.
I've drawn it on rotated graph paper so that we can visualize the ticks along various timelike and spacelike segments.
Instead of 0.97c, I used $v_{Af}=\displaystyle\frac{OC_0}{C_0Z}=\frac{20}{25}=\frac{4}{5}=0.8c$ for convenience. So, $\displaystyle\gamma=\frac{OZ}{C_0Z}=\frac{25}{15}=\frac{5}{3}$.
Note that [from the black segments that are simultaneous to observer-A after observer-A moves] the spatial ordering is preserved up until A's worldline meets B's worldline. So, in the question posed, there is likely some misuse of "length contraction" (which is really about the spatial-separation between two parallel worldlines, e.g. a line parallel to A's worldline after A moves).

A: When A changes frame, it’s now looking at A-B and B-C from that new frame, and they contract when transformed into A’s new frame. B and C remain in their original frame, so there’s no transform, hence A-B and B-C remain the same for them. 
A: Your question is throwing some people off, because you used the words point mass and acceleration. Really, you are considering two distinct point masses whose initial states differ only in their velocities. The inertial reference frames attached to the particles are said to be related by a boost. It is practical to talk about abstract inertial observers instead of concrete point particles located at the origin of inertial frames of reference. It simplifies the language and makes it clear we are only interested in the properties of space-time itself.
Your argument is solid. Accepting the contraction of space due to the Lorentz transformation forces us to conclude that spatial ordering is relative. I really like the setup and notation you have used to demonstrate this. The way you apply special relativity is a little sloppy though. I'll get to that later, but as general advice I would advise you to always reason in terms of events. Events are nice, because in the case of multiple observers it is always clear what is being measured: the event. The observers can simply associate a space-time coordinate to the event, which we can then compare. If we ask multiple observers what the distance between two given observers is, it is not clear at all what we're asking of the observers.
Answer
The standard approach to resolving the paradox is to remember that simultaneity is relative, draw a space-time diagram and show that everything is good and well. I'm already boring myself, so let's try something different.
One of the central themes of special relativity is that space and time stand on equal footing. For every spatial concept, there is a corresponding temporal concept. We have space contraction and time dilation, energy and momentum, density and flux and so on. You have created a spatial paradox. What is the corresponding temporal paradox? I'd like to use your notation, but I'm going to refine your argument a little bit.    
Consider three inertial observers, $A,B$ and $C$. Observers $B$ and $C$ are at
rest with respect to each other and $A$ is not. $A$ observes that $B$ and $C$ are moving towards him and he measures that $B$ passes him (an event) after one unit of time passed on his clock and after having travelled one unit of distance relative to $B$. He passes $C$ after two units of time and two units of distance. In your notation, slightly adapted, the space diagrams are
$$A:A-B-C$$ 
$$B:A-B\ \ |\ \ C$$
$$C:B\ \ |\ \ A--C$$
The vertical bar represent an infinite amount of dashes. From $B$'s perspective, for example, the distance he travels before he meets $C$ is infinite, because they never actually meet. I am forced to flip $A$ and $B$ in the third diagram, because it is impossible to fit an infinite distance ($B\ |\ C$) within a finite distance ($A--C$). Observers $B$ and $C$ disagree. According to $B$, at the beginning of the scenario $A$ is to the left of $B$. $C$ says $A$ starts off at the right of $B$. This is your spatial paradox, slightly reformulated.
The corresponding time diagrams are exactly the same as the space diagrams. The dashes now represent the time intervals that are recorded by the observers. $B$ and $C$ are once again saying something very different. According to $B$, it will meet $A$ in the future. $C$ says $B$ has already met $A$. This is the temporal paradox.
In summary, the scenario starts out as follows. According to $B$, $A$ is on the left of $B$ and will pass $B$ in the future. According to $C$, $A$ is on the right of $B$ and has passed $B$ in the past. These two statements do not contradict each other at all. We see that the spatial paradox has a temporal twin and that they resolve mutually.
Wouldn't it be neat if other paradoxes in special relativity worked like this?
