Happy 2019 (sort of)!

Let the required solution (the smallest multiple of 11 with 0 as the middle digit) be $11n$ . It is obvious that $n > 9$ . Let $10 \le n \le 19$ and let $n = 10+m$ , where $0 \le m \le 9$ . Then we have:

$\begin{aligned} 11n & = 11(10+m) \\ & = 100 + 10 + 10m + m \\ & = 100 + 10(1+m) + m \end{aligned}$

For the middle digit to be 0, $1+m = 10$ $\implies m = 9$ and $11n = 100 + 100 + 9 = \boxed{209}$ .

We seek the smallest three-digit number $p0q$ with alternating sum $p-0+q=11$ . By inspection, we have $\boxed{209}$ . Next up, $308, 407,...,902$ .