Q:

Highly composite numbers

A highly composite number is a positive integer that has more divisors than any smaller positive integer has.  This is OEIS sequence A002182.  Its first 20 terms are
1, 2, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, 5040, 7560

For example, 4 is in the sequence because it has 3 divisors (namely 1, 2, 4), whereas 3 only has 2 divisors, 2 also has 2 divisors, and 1 has 1 divisors. 
Challenge
Given a positive integer input n, output either the n-th highly composite number or the first n highly composite numbers, at your choice (but the choice must be the same for every input n). 
Rules
The program or function should theoretically work for arbitrarily large inputs given infinite time and memory, and without considering data type limitations.  Essentially, this means no hardcoding a finite number of values. 
In practice, the program or function should run in a reasonable amount of time, say less than 1 minute, for n up to 20.  Maximum input or output may be limited by your language standard data type (but again, the algorithm should theoretically work for arbitrarily large numbers). 
Any reasonable input and output format is allowed, including unary. 
Code golf.  Fewest bytes wins. 

A:

Jelly, 15 bytes
,®ÆDL€ṛ©0>/?µƓ#

For input n, this prints the first n highly composite numbers. 
For n = 20, it takes less than two seconds on Try it online! 
How it works
,®ÆDL€ṛ©0>/?µƓ#  Main link.  No implicit input. 

            µ    Push the chain to the left on the local link stack. 
             Ɠ   Read an integer n from STDIN. 
              #  Execute the chain for k = 0, 1, 2, ..., until it returned a truthy
                 value n times.  Return the list of matches. 

,®               Pair k with the value in the register (initially 0). 
  ÆD             Compute the divisors of k and the register value. 
    L€           Count both lists of divisors. 
           ?      If
         >/        k has more divisors than the register value:
      ṛ©             Save k in the register and return k.
        0          Else: Return 0. 

Alternate version, 13 bytes (non-competing)
While the below code worked in the latest version of Jelly that precedes this challenge, the implementation of M was very slow, and it did not comply with the time limit.  This has been fixed. 
ÆDL®;©MḢ’>µƓ#

Try it online! 
How it works
ÆDL®;©MḢ’>µƓ#  Main link.  No implicit input. 

          µ    Push the chain to the left on the local link stack. 
           Ɠ   Read an integer n from STDIN. 
            #  Execute the chain for k = 0, 1, 2, ..., until it returned a truthy
               value n times.  Return the list of matches. 

ÆD             Compute the divisors of k.
  L            Count them. 
   ®;          Append the count to the list in the register (initially 0 / [0]). 
     ©         Save the updated list in the register. 
      M        Obtain all indices the correspond to maximal elements. 
       Ḣ       Retrieve the first result. 
        ’>     Subtract 1 and compare with k.
               This essentially checks if the first maximal index is k + 2, where
               the "plus 2" accounts for two leading zeroes (initial value of the
               register and result for k = 0). 

A:

05AB1E, 15 14 bytes
The input in zero-indexed.  That means that n = 0 gives 1, n = 1 gives 2, etc.  Code:
$µ>DÑgD®›i©¼}\

Explanation:
$               # Pushes 1 and input
 µ              # Counting loop, executes until the counting variable is equal to input
  >             # Increment (n + 1)
   DÑ           # Duplicate and calculate all divisors
     gD         # Get the length of the array and duplicate
       ®        # Retrieve element, standardized to zero
        ›i  }   # If greater than, do...
          ©     #   Copy value into the register
           ¼    #   Increment on the counting variable
             \  # Pop the last element in the stack

Calculates n = 19, which should give 7560 in about 10 seconds. 
Try it online! 
Uses CP-1252 encoding. 

A:

MATL, 26 24 bytes
~XKx`K@@:\~s<?@5MXKx]NG<

Try it online! 
The current largest number of divisors found is kept in clipboard K. Highly composite numbers (HCN) are kept directly on the stack.  A loop keeps testing candidates to HCN.  When one is found it is left on the stack, and clipboard K is updated.  The loop exits when the desired number of HCN have been found. 
~         % take input implicitly (gets stored in clipboard G).  Transform into a 0 
          % by means of logical negation
XKx       % copy that 0 into clipboard K, and delete
`         % do...while
  K       %   push largest number of divisors found up to now
  @       %   push iteration index, i: current candidate to HCN
  @:      %   range [1,...,i]
  \       %   modulo operation
  ~s      %   number of zeros.  This is the number of divisors of current candidate
  <       %   is it larger than previous largest number of divisors? 
  ?        %   if so: a new HCN has been found
    @     %     push that number
    5M    %     push the number of divisors that was found
    XKx   %     update clipboard K, and delete
  ]       %   end if
  N       %   number of elements in stack
  G<      %   is it less than input?  This the loop condition: exit when false
          % end do...while implicitly
          % display all numbers implicitly

