Find pairs of numbers with a particular LCM and GCD
Mathematica, 8 bytes

{4,5092}


Proof that 4 and 5092 are the only solutions: The original problem can be rewritten as 


  x (x + 2010) = 2014 GCD(x, x+2010)2


Let's write x as 2a2 3a3 5a5… and x + 2010 as 2b2 3b3 5b5… Then the equation becomes


  2a2+b2 3a3+b3 5a5+b5… = 2014 22min(a2,b2) 32min(a3,b3) 52min(a5,b5)…


Since 2014 = 2 × 19 × 53, we have


  ap + bp = 2min(ap, bp) + { 1 if p &isin; {2, 19, 53}, 0 else }


Thus


  ap = bp if p ≠ 2, 19, 53
  ap = bp±1 else


Thus


  x + 2010 = 2±1 19±1 53±1 x


There are only 8 possible choices, and we could easily check that 4 and 5092 are the only positive integer solutions. 

Wait, I hear people screaming standard loophole…

Mathematica, 45 bytes

Select[Range[9^7],2014GCD[#,s=#+2010]^2==s#&amp;]