Does the attached mass go up or down in the spring paradox? Take a look at this video.
It shows that when 2 springs in series are transformed into 2 springs in parallel the attached mass goes up. But I'm interested in what happens in that split-second after cutting the blue cord but before the green and red ropes become taut (i.e. when the springs are pulling their respective ropes but the latter are still slack). Does the attached mass go up during that time? Down? Or does it maybe depend on the actual springs used (their pulling strength or something)?
 A: Summary
We do not know if the hanging weight will have a an instant of going downward or not
It depends on the mass and spring constant of the lower spring and mass of the hanging weight.
Logic
We cut the rope. There is gravitational force on the weight. The springs begin to contract rapidly and contract until the strings are tight. (John Hunter’s answer assumes the mass has to descend to make the strings taut, but the contracting springs do most of the moving to tighten the strings).
Assume the strings have no mass.
After cutting but before tight strings, the [lower spring / hanging weight] is a free system not affected by the upper spring or weightless un-taut strings. Let’s consider this system during that time:
The system as a whole is only acted on by gravity so as mentioned by alephzero it will fall - it’s center of mass will accelerate downward at F/M = g, the acceleration of gravity, where M = m_spr + m_hw (spring and hanging weight).
In addition to that, the spring (which has mass) will contract. As alephzero said, if the hanging weight has a much much higher mass than the lower spring, the mass will go down as this system falls (until the strings are tight).
But what if the spring doesn’t have a mass that is negligible compared to the hanging weight? There are cases where the spring constant and/or spring mass are high enough that the mass never (even instantaneously) has a downward acceleration of velocity. The spring contracts with enough inertial force that it’s instantaneous force on the weight is greater than m_hw g.
To be sure there are such cases, take an extreme case where the mass of the hanging weight is negligible compared to the spring’s mass and the spring constant is high. The spring will begin to fall under its own weight while contracting rapidly about its center and its lower end will go up (hence the hanging weight will)
A: Consider the bottom spring and mass as a single system.
When the blue string is cut, there is a net external force on the bottom spring+mass equal to their weight. Therefore the center of mass of the bottom spring+mass will start to move down with acceleration $g$.
However there are also the internal forces in the spring which are no longer balanced by the force from the blue string.
Assuming the mass of the spring is small compared with the weight, this will cause the top of the spring move downwards with an acceleration much bigger than $g$, and the mass to move downwards with an acceleration slightly smaller than $g$.
Similarly the bottom of the top spring will move up, with an acceleration much bigger than $g$.
When the green and/or red strings become tight, the situation changes because of the tension in those strings.
In the experiment where the blue string is cut, all this happens so fast that you can't see the weight's small movement down before it starts to move up. A high speed video would show it, if there was also a fixed vertical scale visible to measure the position of the weight accurately.
A: It's a good question.
The answer seems to be about what happens between 55 and 56 seconds on the video.
At 55 seconds, it shows the red string like the left image below.

But surprisingly the string is only about 0.5cm longer than the right image - you can test this by using a piece of string about 30cm long.
So, the weight does drop, but only by about 0.5cm until the string becomes tight and bounces the weight back up.
The video doesn't show it, (it seems only possible to stop it at 1 second intervals), but how long would it take?
Using the equation of motion $$s=\frac{gt^2}{2}$$ where $s$ is the drop distance  of 0.5cm gives a time of $0.03$ seconds, and the video simply doesn't show it, even frame by frame.
A: When the blue cord is cut, several things happen very rapidly.  At the instant the blue cord is cut, the two springs in series become unloaded, and they rapidly contract up until the red and green cords become taut.  At that point, the springs become loaded again and their rate of contraction would change in a complicated way based on what the weight does.
While the springs are contracting and before the red and blue cords become taut, the weight is free falling, so it is moving down under the acceleration of gravity.  When the red and blue cords become taut, the springs "catch" the weight that is falling.  Because the time interval for this catch is so short, the weight has a very low downward velocity, and this velocity must still be stopped and reversed to pull the weight upward.  Because the weight has inertia, this means that for an instant, the upward contraction of the springs is reversed by a very small amount.  After that instant, the two springs will continue contracting upward at a rate that is somewhat slower than when the weight was in free fall.
Note that it would take high speed video (e.g., 100,000 frames per second) to observe these changes in the cords, springs, and weight, so these changes are not observable in the posted video.
A: This is an excellent question, and its real answer is all about how the springs contract.
As a matter of fact, to answer this question, we cannot think about the springs in an idealized way, we must consider that the springs are extended objects with mass and tension. As such, when the blue rope is cut, the relaxation of the springs happen in the form of a relaxation shock wave that takes time to travel from one end of the spring to the other. This relaxation shock wave has the property that the spring coils in front of it remain perfectly taught until they are reached by the shock front. To these parts of the spring it is as if the rope had not been cut at all. As the shock front passes a spring coil, it fully relaxes, turning its tension energy into kinetic energy. Once it is fully relaxed, it's moving at constant speed along with the rest of the already relaxed spring coils. The process looks a bit like this:
\/\/\/\/\/\/\/\/\/\/ fully extended, no movement
\/\/\/\/\/\/\/\/\/V
\/\/\/\/\/\/\/\/VV   relaxation shock wave traveling left
\/\/\/\/\/\/\/VVV
\/\/\/\/\/\/VVVV
\/\/\/\/\/VVVVV
\/\/\/\/VVVVVV
\/\/\/VVVVVVV
\/\/VVVVVVVV
\/VVVVVVVVV
VVVVVVVVVV          fully relaxed, moving left
NVVVVVVVV
NNVVVVVV            compression shock wave traveling right
NNNVVVV
NNNNVV
NNNNN               fully compressed, not moving

The key insight is, that while the springs relax and become shorter, the far end of the springs remain utterly ignorant of the relaxation wave that's moving closer. The consequence is, that the long ropes become taught before the weight at the far end of the lower spring realizes what's going on.
With these insight, we can now fully describe what will be happening:

*

*When the rope is cut, the weight remains suspended beneath the lower spring.


*As the relaxation shock wave moves up the upper spring, the red rope becomes taught. The added force along the red rope causes the weight to accelerate upwards.


*The sudden onset of force from the red and green ropes causes a second shock wave to start traveling across the springs. This time, the shock wave stops the movement of the springs, turning their kinetic energy back into tension.


*On each spring, a wave of a relaxed portion is moving towards the outer ends, both ropes are taught with enough tension to hold the weight by themselves, and the weight is accelerating upwards at $1g$.


*The relaxation shock waves hit the ends of the springs and reflect off it, becoming compression shock waves in the process.


*While the compression shock wave moves up the lower spring and the following tension wave has not reached its lower end, the lower spring pushes the weight downwards. This cancels the force from the red rope, making the weight enter free fall.


*The quickly following tension shock wave restores the force on the weight, allowing it to continue its acceleration upwards.


*As the waves travel up and down the springs, they become less sharp and dissipate their excess energy, making the springs more smoothly tensioned. At the same time, the weight enters an oscillation around its final position.


*All these oscillations dissipate their remaining energy and die away until the system enters its new equilibrium.
TL;DR:
Due to the retardation of the signal about the cut wire along the lower spring, the weight first gets the signal of the taught red rope and starts its movement with accelerating upwards.
A: In the video he shows what happens when he releases the blue rope slowly. You can see that until the red and green cords become taut, the springs are still in series and the weight goes down.
