Recall the rules for divisibility by $9$ and $11$:

• A number is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Since the number has only $1$'s, it must have a multiple of $9$ number of digits. 
• A number is divisible by $11$ if and only if the alternating sum of its digits is divisible by $11$. Since the number has only $1$'s, the alternating sum of its digits will be $1$ if it has an odd number of digits, or $0$ if it has an even number of digits. Thus, we need the number to have an even number of digits.

Since we need the number to have a number of digits that is a multiple of $9$ and even, the smallest possible number of digits it may have is $18$.