***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

<hr>
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); }