Who likes 11 so much?

I ain't certain that my solution is error-free, but that's what I did:-

$11^k=(10+1)^k$ , and $100|11^k-1$ means $100|(10+1)^k-1$ . Now, all terms in the expansion of $(10+1)^k$ will have $100$ in it except the last and the second last terms, which are $1$ and $10$ respectively. So, opening the brackets and removing all the multiples of $100$ and canceling positive and negative ones, we get $100|10k$ , or $k=10a$ , where $a$ is a natural number less than $100$ . So sum of all possible values of $k$ is $10+20+30+............+980+990$ , which is an AP summing up to $\boxed{49500}$ .

${ 11 }^{ 10 }\equiv 1\quad (mod\quad 100)$ . Hence the values of k are 10, 20, 30 ,..............., 980, 990.
Sum of all these values equals to $\boxed {49500}$ .

We can use Pascal's triangle to check for powers of 11: http://www.mathsisfun.com/pascals-triangle.html

(I am supplying this link because I found it troublesome to print the Pascal's triangle here).

If we actually check the triangle, we find that the last two digits of each successive power (starting from 11) are: 11,21,31.41.51,.... If we go on, we find that in the 10th row, the last two digits are 01. So, we conclude

${ 11 }^{ 10 }\equiv 1(mod\quad 100)$ . (A)

Now we use the property that if $a\equiv b(mod\quad m)$ , then ${ a }^{ k }\equiv { b }^{ k }(mod\quad m)$ in equation (A), to find that

${ 11 }^{ k }\equiv 1(mod\quad 100)$ , for k=10,20,30,......,990.

the sum of all these values is

$10\times \frac { n(n+1) }{ 2 }$ , where n=99. Therefore, we get the answer as 49500.

The rest of 11^k/100 must be the same as 1/100. That said, any power of 11 that ends with X01 is a valid solution for our problem. Analysing the behavior of powers of 11 you will notice that the powers 10,20,30... follow the pattern that we wanted. From that point, we will need from 11^10 to 11^990 (notice the "last than 1000" in the question) giving us a sum of 10(1+2+3+4+...+99) = 10 49.5 100 = 49500.