We call a number *eligible* if its digits are all either $1$ or $2$. A number is divisible by $99 = 9 \cdot 11$ if and only if it is divisible by both $9$ and $11$. Suppose $N \in \mathbb{N}$ has base-10 expansion $a_n a_{n-1} \dots a_2 a_1$. We define three digit-sums (full, odd, even):
$$S(N) := \sum_{1 \le i \le n} a_i, \quad o(N) := \sum_{\substack{1 \le i \le n \\ i \text{ odd}}} a_i, \quad e(N) := \sum_{\substack{1 \le i \le n \\ i \text{ even}}} a_i.$$

As is well known (and easily established), $N$ is divisible by $9$ if and only if $S(N)$ is divisible by $9$, while $N$ is divisible by $11$ if and only if $d(N) := o(N) - e(N)$ is divisible by $11$.

Suppose first that $S(N) = 9$. Then $N$ has at least five digits,
$$3 \le o(N) \le 9 - 2 = 7, \quad 2 \le e(N) \le 9 - 3 = 6,$$
and $o(N)$, $e(N)$ are of opposite parity. Consequently, $0 < |d(N)| \le 5$, and so $N$ cannot be divisible by $11$.

Thus, we must have $S(N) \ge 18$ for eligible $N$ to be divisible by $99$. Suppose next that $S(N) = 18$. To minimize eligible $N$ with $S(N) = 18$, we must certainly minimize the number of digits in $N$. We immediately rule out nine $2$s, since then $d(N) = 10 - 8$ is not divisible by $11$.

The next smallest number of digits involves picking eight $2$s and two $1$s. The smallest such number is the one with $1$s in the leading positions, i.e. $N = 1122222222$. Then $o(N) = e(N) = 9$, and $N$ is divisible by $11$, and hence by $99$. Thus, this is the minimal example with $S(N) = 18$.

It remains only to consider eligible numbers $N$ with $S(N) \ge 27$. Such numbers have at least $14$ digits, so are larger than the one we found above. Thus, the minimal number is indeed $1122222222$.