Find the least number $x$ such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$. Find the least number $x$  such that $ 11$ divides $x$ and sum of its digits $S(x)$ is $27$.
Since $S(999)=27 $ it is clear that the number of digits $n>3.$ Let $x_i$ be digits then we have two equations
\begin{cases}
x_1+x_2+\cdots+x_n=27=5 \mod 11,\\
x_1+x_3+x_5+\cdots=x_2+x_4+\cdots \mod 11.
\end{cases}
It follows that 
$2(x_2+x_4+\cdots+)=5 \mod 11$ 
or 
$$
x_2+x_4+\cdots = 8 \mod 11.
$$
For the case $n=5$ it reduced to $x_2+x_4=8$ and I manage to pick up the solution $10989.$ 
Is there better solution?

Since $\sum x_i=27$, $\sum (-1)^{i-1}x_i=27-2(x_2+x_4+\cdots)$ must be odd and divisible by $11$, and hence either $11$ or $-11$ (it must be between $-27$ and $27$.)
If $\sum (-1)^{i-1}x_i=11$ then $x_1+x_3+\cdots  = 19, x_2+x_4+\cdots = 8$. The only way to get $19$ is with three digits, $(x_1,x_3,x_5)=(1,9,9)$ yielding the smallest values in lexicographical order. Then $x_2+x_4=8$ has $(0,8)$ for the smallest lexicographical order. 
So you get $10989$.
If $\sum (-1)^{i-1}x_i=-11$, then $x_2+x_4+\cdots = 19$ and thus that there must be at least $6$ digits.
