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?

When $n = 5$, $10989$ is already the least solution to the problem.
First, when $n \le 4$, no solution exists.  For $n = 3$, we deduce from the sum of digits $S(x) = 27$ that $x = 999$, which is not a multiple of $11$.  For $n=4$, we have $x_2 + x_4 \equiv 8 \pmod{11}$.
Since each digit is an integer from $0$ to $9$ inclusive, we have $$0 \le x_n \le 9 \forall n \in \Bbb N^*,$$
from which we deduce that $x_2 + x_4 = 8$.  Due to the sum of digits $S(x) = 27$, we have $x_1 + x_3 = 19 > 18$, which is clearly false.
We conclude that no solutions exist when $n \le 4$.
Now we come back to $n = 5$.  We're given the number $10989$ which satisfies the given conditions, and we hope to prove that it's the minimal solution to the problem.  From the condition in the OP $$x_2+x_4 (+\cdots) = 8 \pmod{11}.$$ and by observing the second and fourth digit, $0$ and $8$ are already the best combination.  Since it's a five digit number, the leftmost digit is at least one.  So we're left with two digits.  By considering the sum of digits $S(x) = 27$, we know that the sum of the remaining two digits are $27 - 1 - 8 = 18$.  This forces us to choose $9$ for the remaining two digits.  Therefore, $10989$ is already the least natural number satisfying the given conditions in the OP.
