4 digit numbers divisible by 11 Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places. But how do I use this to get the answer? 

Let $[abcd]$ be the number. Then $[abcd] \equiv [ab]+[cd] \pmod{99}$ which implies $[abcd] \equiv [ab]+[cd] \pmod{11}$. Because $a,b,c,d \in \{1,2,3,4\}$,
$[abcd] \equiv [(a+c)(b+d)] \pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
\begin{array}{|r|r|r|}
\text{sum} &\text{ac and bd events} & \text{counts} \\
\hline
  2 &                   11 & 1\\
  3 &             12, \ 21 & 4\\
  4 &       13, \ 22, \ 31 & 9 \\
  5 & 14, \ 23, \ 32, \ 41 & 16\\
  6 &       24, \ 33, \ 42 & 9\\
  7 &              34, \ 43 & 4\\
  8 &                    44 & 1 \\
\hline
\text{total} && 44
\end{array}
