Jelly, 11 bytes

sZµṢŒrUṀṪµ€

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Wasn't expecting to beat Dennis on this one, so tried to FGITW it (after trying several possibilities; there's more than one way to make 11). I came in shorter, much to my surprise.

Takes the string then the count as command-line arguments. Outputs on stdout.

Explanation

sZµṢŒrUṀṪµ€
s            Split {the first input} into {the second input}-sized groups
 Z           Transpose
  µ      µ€  On each of the transposed groups:
   Ṣ           Sort it;
    Œr         Run-length encode it;
      U        Rearrange it to the form {count, letter};
       Ṁ       Take the largest element (i.e. largest count)
        Ṫ      Take the second element of the pair (i.e. just the letter)

This uses the insight that the letter in each position of the pattern must be the most common letter corresponding to that position. We can find the letters corresponding to a particular pattern via splitting into pattern-sized groups, and transposing. The main reason this solution is so long is that Jelly doesn't seem to have a short way to find the mode of a list (I made several attempts, but they're all at least six bytes long).

Jelly, 10 bytes, based on @Dennis' solution

⁸ċ$ÞṪ
sZÇ€

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This is a combination of @Dennis' solution and my own; there was a five-byte mode in that solution, which I stole for this solution. (I already had solutions based on ⁸ċ, but couldn't get below six bytes with it; I hadn't thought of using Þ.)

Explanation

µ…µ€ and Ç€ (with the … on the previous line) are both three bytes long (the latter needs a newline), and equivalent. Normally I use the former, but the latter's more flexible, as it allows you to use ⁸ to mention the argument.

This makes it possible to sort (Þ) by the number of occurrences in ⁸ (⁸ċ), then take the last element (Ṫ), to find the mode in just five characters.

Mathematica, 41 bytes

#&@@@Commonest/@(#2~Partition~UpTo@#)&

Input and output are lists of characters.

Also based on the modes of the lines of the transpose. I believe they called the built-in for the mode of a list Commonest solely to spite code golfers.

For only three additional bytes and without any restructuring, we can get a list of all optimal solutions:

Tuples[Commonest/@(#2~Partition~UpTo@#)]&

Python 2, 106

Now it's a different answer! I was thinking about one(almost)-liner from the beggining. Now even shorter, based on zip usage by @Rod.

Thanks to @L3viathan and @Rod for clarification about using lambdas as answer

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lambda S,N:max(combinations(S,N),key=lambda s:sum(x==y for x,y in zip(S,s*len(S))))
from itertools import*

Explanation:

combinations(S,N) creates all combinations length N from characters of S

max() have argument key which takes as input function to use to compare elements

lambda s:sum(x==y for x,y in zip(S,s*len(S))) passed as such function

This lambda counts number of matching characters in list of tuples, produces by zip(S,s*len(S))

s - one of combinations and it's multipled by len(S) which creates string that is guaranteed longer than S

zip creates tuples of characters of each string S and s*len(S) and ignores all characters that can't be matched (in case of one string longer than another)

So max chooses combination, that produce maximum sum

Python 3, 99, 73 61 bytes

-12, thx to @Rod

lambda s,n:''.join(max(s,key=s[i::n].count)for i in range(n))

Same idea, but rewrote it to eliminate the import statement.

lambda s,n:''.join(max(s,key=lambda c:s[i::n].count(c))for i in range(n))

Original

from collections import*
lambda s,n:''.join(Counter(s[i::n]).most_common(1)[0][0]for i in range(n))

Explanation:

s[i::n]                  a slice of every nth character of s, starting at position i

Counter(s[i::n])         counts the characters in the slice
  .most_common()         returns a list of (character, count) pairs, sorted by decreasing count
    [0][0]               grabs the letter from the first pair (i.e., the most common letter
      for i in range(n)  repeat for all starting positions

''.join                  combines the most common letters into a single string

Jelly, 12 11 bytes

s@ZċþZMḢ$€ị

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How it works

s@ZċþZMḢ$€ị  Main link. Arguments: n (integer), s (string)

s@           Split swapped; split s into chunks of length n.
  Z          Zip/transpose, grouping characters that correspond to repetitions.
   ċþ        Count table; for each slice in the previous result, and each character
             in s, count the occurrences of the character in the group.
             This groups by character.
     Z       Zip/transpose to group by slice.
        $€   Map the two-link chain to the left over the groups.
      M        Find all maximal indices.
       Ḣ       Head; pick the first.
          ị  Index into s to retrieve the corresponding characters.

JavaScript (ES6), 105 bytes

(n,s)=>[...Array(n)].map((_,i)=>[...s].map((c,j)=>j%n-i||(o[c]=-~o[c])>m&&(m++,r=c),m=0,o={})&&r).join``

Pyth, 11 bytes

meo/dNd.TcF

Takes input as s,n and outputs as a list of characters.

Explanation

meo/dNd.TcF
         cFQ   Split s into chunks of length n.
       .T      Transpose.
m o/dNd        Sort characters in each string by frequency.
 e             Take the most common.

Japt, 16 15 bytes

Saved 1 byte thanks to @obarakon

Ç=VëUZ)¬ñ!èZ o

14 bytes of code + 1 byte for the -P flag. Try it online!

Ungolfed and explanation

 Ç   =VëUZ)¬ ñ!èZ o
UoZ{Z=VëUZ)q ñ!èZ o}
                          Implicit: U = input number, V = input string
Uo                        Create the range [0...U).
  Z{               }      Map each item Z by this function:
      VëUZ                  Take every U'th char of V, starting at index Z.
    Z=    )                 Call the result Z.
           q                Split the result into chars.
             ñ!èZ           Sort each char X by the number of occurrences of X in Z.
                  o         Pop; grab the last item (the most common char).
                      -P  Join the results (array of most common chars) into a string.

Python 2, 132 bytes

from itertools import*
p,k=input()
b=l=len(p)
for i in combinations(p,k):
 x=sum(x!=y for x,y in zip(p,i*l))
 if x<b:b,o=x,i
print o

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05AB1E, 17 bytes

Iôð«øvy{.¡é®èÙJðÜ

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Explanation

Iô                 # split 2nd input in chunks of 1st input size
  ð«               # append a space to each
    ø              # zip
     vy            # for each y in the zipped list
       {           # sort the string
        .¡         # group into chunks of consecutive equal elements
          é        # sort by length
           ®è      # pop the last element (the longest)
             Ù     # remove duplicate characters from the string
              J    # join the stack into one string
               ðÜ  # remove any trailing spaces

PHP, 245 Bytes

function p($c,$s,$r=""){global$a;if(($c-strlen($r)))foreach(str_split(count_chars($s,3))as$l)p($c,$s,$r.$l);else{for($v=str_pad("",$w=strlen($s),$r);$z<$w;)$t+=$v[$z]==$s[$z++];$a[$t][]=$r;}}p($argv[1],$argv[2]);ksort($a);echo join(" ",end($a));

Online Version

Breakdown

function p($c,$s,$r=""){
    global$a;
    if(($c-strlen($r)))  # make permutation
        foreach(str_split(count_chars($s,3))as$l)
            p($c,$s,$r.$l); #recursive
    else{
        for($v=str_pad("",$w=strlen($s),$r);$z<$w;) 
        $t+=$v[$z]==$s[$z++]; #compare strings
        $a[$t][]=$r; # insert value in array
    }
}
p($argv[1],$argv[2]); #start function with the input parameter
ksort($a); # sort result array 
echo join(" ",end($a)); #Output