So Last Year

To begin with, let us notice that the number $1000^{2014}$ is in fact a $1$ followed by a lot of $0$ 's. In particular, it has $2014\cdot 3=6042$ zeros in total.

Now let us also notice that any number in the form of $10^{b}$ , where $b\geq 5$ gives a somewhat similar result when you subtract $2014$ from it.

Let us look at a few examples:

$100000-2014=97986 ;$ $1000000-2014=997986 ;$ $10000000-2014=9997986$

Additionally, it always ends in the sequence $7986$ and has an $N$ number of $9$ 's in front of it, where $N$ is the number of digits of $10^{b}$ minus $5$ (we lose $1$ digit due to the subtraction and $4$ are reserved for the ending sequence).

Now let us take a look at our problem again. $1000^{2014}$ is indeed a number of the sort mentioned above, and thus we may apply the rules to it. Since it is a $6043$ digit number, we know that it will have $6038$ $9$ 's and it will also end in $7986$ when you subtract $2014$ from it.

Finally, the sum of the digits of $1000^{2014}-2014$ is $6038\cdot 9 +7+9+8+6=\boxed{54372}$

1000^2014 has 2014 * 3 '0's . Now if we subtract 2014 from that number. We get 999999999 . . . 999997986. Number of 9s in this number =(3 * 2014)-4. Here 4 is subtracted as that the last four digits are different. Now adding 7+9+8+6+((3 2014)-4) 9) = 54372