For how many whole numbers n is n added to the sum of its own digits equal to 2016? (For instance 1234 added to 1 + 2 + 3 + 4 gives 1244, not 2016.)

A four-digit $n$ below 2016 has a digit sum of at most 1 + 9 + 9 + 9 = 28, so $n$ cannot be smaller than 2016 − 28 = 1988; only the range 1988 to 2016 need be searched. In the 1990s write $n=1900+10a+b$, so $n+S(n)=1900+10a+b+(10+a+b)=1910+11a+2b=2016$, giving $11a+2b=106$, whose only digit solution is $a=8,b=9$ — that is $n=1989$, and indeed 1989 + 27 = 2016. In the 2000s write $n=2000+10a+b$, so $n+S(n)=2002+11a+2b=2016$, giving $11a+2b=14$ and hence $a=0,b=7$ — that is $n=2007$, and 2007 + 9 = 2016. Both branches are exhausted, so exactly 2 numbers work. Trap 1 finds the easier 2007 and stops without checking the previous decade. Trap 0 assumes none can work because 2016 itself fails. Trap 3 throws in 1998 as well, but 1998 + 27 = 2025.