The sum of the digits of a multiple of $3$ is divisible by $3$. So, we are looking for numbers with the sum of the digits divisible by $21$. The sum of the digits of any number smaller than $1000$ is less than or equal to $9 + 9 + 9 = 27$. So, in our case the sum of the digits is equal to $21$.

The smallest possible digit is at least $21 - 18 = 3$ and at most $\frac{21}{3} = 7$. The only possible triples of digits are $(3, 9, 9)$, $(4, 8, 9)$, $(5, 7, 9)$, $(5, 8, 8)$, $(6, 6, 9)$, $(6, 7, 8)$ and $(7, 7, 7)$. Each triple with three different digits corresponds to six different numbers. Each triple with two equal digits corresponds to three different numbers and the triple $(7, 7, 7)$ corresponds to one number. Hence, there are $3 \cdot 6 + 3 \cdot 3 + 1 = 28$ numbers with the required properties.