Ex-Increasing Set Sequence
Brachylog, 28 bytes

⟦₁⊇ᶠk⊇pSs₂ᶠ{c≠&amp;⟨+/l⟩ᵐ&lt;ᵈ}ᵐ∧Sl


Try it online!

This is really damn slow. Takes about 30 seconds for N = 3, and it didn't complete after 12 minutes for N = 4. 

Explanation

⟦₁                             Take the range [1, …, Input]
  ⊇ᶠk                          Find all ordered subsets of that range, minus the empty set
     ⊇                         Take an ordered subset of these subsets
      pS                       Take a permutation of that subset and call it S
       Ss₂ᶠ                    Find all substrings of 2 consecutive elements in S
           {           }ᵐ      Map for each of these substrings:
            c≠                   All elements from both sets must be different
              &amp;⟨+/l⟩ᵐ            And the average of both sets (⟨xyz⟩ is a fork like in APL)
                     &lt;ᵈ          Must be in strictly increasing order
                         ∧Sl   If all of this succeeds, the output is the length of L.


Faster version, 39 bytes

⟦₁⊇ᶠk⊇{⟨+/l⟩/₁/₁}ᵒSs₂ᶠ{c≠&amp;⟨+/l⟩ᵐ&lt;₁}ᵐ∧Sl


This takes about 50 seconds on my computer for N = 4.

This is the same program except we sort the subset of subsets by average instead of taking a random permutation. So we use {⟨+/l⟩/₁/₁}ᵒ instead of p.

{         }ᵒ     Order by:
 ⟨+/l⟩             Average (fork of sum-divide-length)
      /₁/₁         Invert the average twice; this is used to get a float average


We need to get a float average because I just discovered a ridiculous bug in which floats and integers don't compare by value but by type with ordering predicates (this is also why I use &lt;ᵈ and not &lt;₁ to compare both averages; the latter would require that double inversion trick to work).