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This is a comment at the comments of Gerhard "Still Computing Oh So Slowly" Paseman's answer, giving just indexes n for more record-holders.

I used the function fharm(n,eps) = frac(h(n))*n^(1+eps) to find the local minima for a fixed eps and consecutive arguments n. The n-1 of my list should agree with the n in the comments. (The harmonic numbers can be computed in Pari/GP using h(n) = psi(1+n) + Euler )

• Seek manually the first n having a local minimum for w=fharm(n,eps) and document n,w.
• Guess the next n by n=ceil(n*exp(1))+10 compute w and decrement n as long as also w=fharm(n,eps) decreases .

This is what I got using Pari/GP (and 800 digits internal precision, don't know exactly what it needs for the psi-function at this arguments n) in a couple of seconds:

               n     |   fharm(n,0.1) 
 --------------------+----------------        
                   1    0.0 
                  11    0.277901138881
                  31    1.19066007386
                  83    0.267050685436
                 227    1.70522868276
                 616    1.49190552481
                1674    1.70759404228
                4550    2.19789655729
               12367    1.36471771489
               33617    1.68810700160
               91380    0.873923093872
              248397    1.05774925083
              675214    3.52030034776
             1835421    2.93703145896
             4989191    2.22651939403
            13562027    1.04059409912
            36865412    0.783097603793
           100210581    6.15552338765
           272400600    3.07296513521
           740461601    2.29064626275
          2012783315    2.37389093860
          5471312310    1.01950811681
         14872568831    3.51618209965
         40427833596    2.82622840722
        109894245429   10.6250101261
        298723530401    7.64454034138
        812014744422    6.96806150688
       2207284924203    3.81951285918
       6000022499693    2.46795156044
      16309752131262   12.4771965474
      44334502845080    1.85826500223
     120513673457548   14.1650514186
     327590128640500   21.4949794587
     890482293866031    0.75550636057
    2420581837980561   21.3699821691
    6579823624480555    3.02859881920
   17885814992891026    3.21976135398
   48618685882356024    1.64093320527
  132159290357566703   13.7707827691
  359246197441016284   14.6487253696

The set $M$ is here the set of all $n$ where fharm(n,eps) < 1 . The number of such entries for n=359246197441016284 ~ 10^17 is here 6 . The logarithm of the indexes nare

    ln(n)         ln(n) + gamma
  -------------------------------
        0.E-809  0.577215664902
  2.39789527280   2.97511093770
  4.41884060780   4.99605627270
  11.4227819151   11.9999975800
  17.4227843253   17.9999999902
  34.4227843351   35.0000000000

This - and the differences - might be interesting for extrapolating a tendency and a guess for the cardinality of $M$ depending on eps and the range of n being checked.
Anyway, all numerical test which I've done indicate that the cardinality of $M$ is finite and even decreases with increasing $\varepsilon$ and only if $\varepsilon=0$ might be infinite.

---

This is the Pari/GP-program which I used. Note, that the quotient of two consecutive $n$ approaches $\small \exp(1)$ so I need only the following guess:

Using the function

  fharm(n,epsilon=0.1) = frac(psi(1+n)+Euler) * n^(1+epsilon) 

the following needs only a couple of steps to find the n with the next local minimum when called with n having the current local minimum:

{findnext_n(n,eps=0.1)=local(n1,w1,n2,w2);
 n1=ceil(n*exp(1)+10); w1 = fharm(n1,eps);
 n2=n1-1;              w2 = fharm(n2,eps);
 while( w2<w1,   n1=n2;w1=w2; 
                 n2=n1-1;w2=fharm(n2,eps)
      );
 return(n1);}               
         

So it is applied:

 \\ test it --------------------
 eps=0.1
 {listlen=40;
 list=matrix(listlen,2);
 list[1,]=[n=1,fharm(n,eps)];
 for(k=2,listlen,
       n=findnext_n(n,eps);
       list[k,]=[n,fharm(n,eps)];
      );
 printp(list); }