He was wrong too

I have the book "Wheels, Life, and other Mathematical Amusements" (which is the one referenced by your links), the 1996 edition. In this book Gardner only gives the suboptimal solution with 20 cubes (7+7+2+2+1+1).

Okay, finally I'm back in town, and I think maybe there does exist a solution after all. A 1+5+6+5+6+1=24 configuration. It's 1:00 am right now and I'm going to sleep. In the morning, I'll see if I can put together a graphic to see if it's a real solution.

Edit: I have 2 critical lines to match in order to avoid an interference. In spite of working on it all morning, I can't seem to close the gap. This is the kind of thing that looks good when you draw a sketch, but computer generated reality shows you something else.

Take a look at the animation I made below, to see how it all comes together.

Nope. 22 does work. try with your own ice cubes.

Sorry, I don't really understand what you are pinpointing to. But if on each of the 6 faces, all 4 red cubes touch the center point of a side of the blue cube, the cubes will intersect each other.

The face (excluding corners and edges) of a red cube needs to touch the face (excluding corners and edges) of a blue cube.

Otherwise, we can have the standard 26 cube arrangement around a central cube.

There is no requirement for the blue cube to be completely covered from view.

The only requirement is that the red cubes touche the blue cube.

Please provide a visual illustration. I've reached the same answer as Viktor. and I think vertical variance in the middle layer will eat away from the top/bottom layers.

Since no red cube can simultaneously be in face contact with the blue cube in more than one place, any solution can be expressed as how the red cubes are partitioned. For example, the solutions offered here can be expressed as 7+1+2+1+2+7=20, which is not optimum. I know of a 1+5+5+5+5+1=22 solution, which is clear cut. Without actually giving away how the "correct" solution looks like, perhaps how it's partitioned can be revealed, and give others the chance to figure out the final form?

A 8+2+2+2+2+8=24 solution seems incredibly out of reach, and a 7+2+3+2+3+7=24 solution seems far out of reach as well. However, a 7+4+4+4+4+1=24 solution seems possible, while a 4+4+4+4+4+4=24 solution seems even more possible and less restrictive, but I've not been able to find any that works without interference somewhere.

It shouldn't have to be said that any solution cannot exceed 26, which is the kissing number for identical cubes. So a solution of 24 red cubes seems quite remarkable, if only face-to-face contacts are allowed.

I can see how a 7+1+2+1+5+7=23 configuration, a variation of the 7+1+2+1+2+7=20 solution, can work, but I don't see how he squeezes in that extra cube to make it a 7+1+2+1+5+1+7=24 solution. Or is he including the center cube? I'll come back to this later.

I've finally got a peek at Martin Gardner's "Wheels, Life, and Other Mathematical Amusements" from which @Alex Li cites the purported 24 cube solution . Having seen it, I must conclude: 1) The passage provides examples of another 22 cube solution as well as a 23 cube solution that has me convinced. 2) These are given as solutions; even those submitted to him were submitted as solutions, and not as proofs. 3) Any reader who wishes to carefully account his conclusions should see that calling this "a 24 cube solution" is an editorial faux pas... The cube faces he counts provide a configuration of 7+7+1+2+1+5, yet across a split line (possibly as copied from its original submitters) he writes it as 7+7+1+1+2+1+5, and we all know that cubes have only six faces. Carefully accounting the touches he tallies prove a miscount: the solution has only ever provided 23 cubes.

How does it work? I'll try to summarize it with our red & blue colors, and note some limitations & inequalities he glosses past. I'll also use compass directions, hoping that aids visualization.

Start with the blue cube resting north against our first red cube. The second red cube will connect to the east face of the first. These two will always pivot on the corner surface of the blue's SE corner. They surface touch on a significant, but almost infinitesimally small, surface of blue. Now let's open "the mouth" between the SW corner of blue and NW of red by a tiny amount, say less than ten degrees, (Note, if it's opened beyond 15 degrees, the model falls apart before we get to the really interesting part.)

Another pair of reds will pivot on the blue's NE corner, and another pair on blue's NW corner. Notice that the first pairs displace the subsequent ones, so our original ten degree opening at blue's SW forces an opening at least twice that at blue's SE, which gets doubled-or-more again at blue's NE. This final angle, let's call it 'e,' is equal to the angle of the last/sixth red cube overhanging the blue cube's west face like a ballcap bill. We want to minimize this angle.

Since the mouth at blue's SW can be reduced to an arbitrarily small but positive amount, the subsequent angles can likewise be downsized to small but positive amounts, even though the final angle will be at least four times the size of the mouth. In the final configuration, the mouth may need to be as small as a single degree, depending on how infinitesimally small our pivot areas touch, and how narrow we need to reduce the ballcap bill's overhang.

Now that we have six red cubes curled around the back three corners of the blue, we're almost done with the top and bottom layers. Raising the blue cube one side unit to the second plane, we drop in a red cube, and rotate it against the first (southern) red cube, toward the east, the back of the mouth's acute angle. It's a tight fit, but whatever tiny area attaches the pair in the NE corner should only be a minor displacement. So long as that tiny area is not excessive, the seventh red's NW corner should match against the West face of the blue.

Layer three should match the angles and orientation of layer one. Now we have the blue cube alone on layer two, touching seven cubes on the layer below and seven more on the layer above. We're nearly there. Into layer two, slide one cube squarely against the north face, another against the south face, and two more straddling the east face.

Finally, we cover the blue's west face with the five cubes as configured in @Michael Mendrin 's 1+5+5+5+5+1 solution... but instead of the edge-limit-line being oriented to our south, it is rotated 180 degrees to come north against the ballcap bill.

This is why it is a difficult solution: There is a positive margin for five cubes to work at a height that is slightly smaller than the cube's height. The ballcap bill's overhang is reduced to an arbitrarily small, yet positive distance. Since that distance is within the margin of the five cubes, they can all touch the blue's west face without overlapping any other cubes. So the final configuration is 7+7+1+2+1+5=23.

*EDITED 2016-08-12: In the fourth paragraph, I've changed "last/seventh" to now read "last/sixth."

$7+6+4+4+2+2$ There was hardly any symmetry in my $25$ arrangement and so I made an error visualising it.