Denote the sum of the digits of a positive integer $m$ by $S(m)$. Then $S(11) = 2$. We notice that for the first few multiples of $11$ the sum of the digits is always even. The first multiple for which the sum of the digits is odd is $209 = 19 \cdot 11$ and this sum is $11$. Using these two examples we can construct multiples of $11$ with the sum of the digits equal to $n$ for almost all positive integers $n$.

The number of the form $11\dots11$ with $11$ repeated $k$ times in a row is divisible by $11$ and the sum of the digits is $2k$. Hence, for all even $n$ there exists a multiple of $11$ with the sum of its digits equal to $n$.

As we have just seen we can get the sum of $11$ from $209$. Any number of the form $20911\dots11$ with $209$ followed by $k$ repetitions of $11$ is divisible by $11$ and the sum of its digits is $11 + 2k$. So, for all odd integers $n$ greater than or equal to $11$, there exists a multiple of $11$ with the sum of the digits equal to $n$.

The only positive integers still in question are $1$, $3$, $5$, $7$ and $9$. Let us show that in these cases we cannot find multiples of $11$ with the required sum of digits.

Let $m$ be a multiple of $11$. Assume that $S(m) \le 9$. Let $L$ denote the sum of the digits of $m$ in the odd positions and let $D$ be the sum of the digits in the even positions. Since $S(m) = L + D$, we have $L, D \le 9$. The criterion for divisibility by $11$ implies that $L - D$ is divisible by $11$, but on the other hand we have $-9 \le -D \le L - D \le L \le 9$. So, $L - D = 0$ or $L = D$. We conclude that $S(m) = L + D = 2D$ is even and no multiple of $11$ can have the sum of its digits equal to $1$, $3$, $5$, $7$ or $9$.