Not quite right, since $9\times 0 = 0$ and the digits don't add up to $9$; but otherwise correct.
The reason it works is that we write numbers in base $10$, and when you divide $10$ by $9$, the remainder is $1$. Take a number, say, $184631$ (I just made it up). Remember what that really means:
$$184631 = 1 + 3\times 10 + 6\times 10^2 + 4\times 10^3 + 8\times 10^4 + 1\times 10^5.$$
The remainder when you divide any power of $10$ by $9$ is again just $1$, so adding the digits gives you a number that has the same remainder when dividing by $9$ as the original number does. Keep doing it until you get down to a single digit and you get the remainder of the original number when you divide by $9$, except that you get $9$ instead of $0$ if the number is a nonzero multiple of $9$.
But since every multiple of $9$ is a multiple of $9$, you will always get $9$.
Note that you have a similar phenomenon with $3$ (a divisor of $9$), since adding the digits of a multiple of $3$ will always result in one of the one-digit multiples of $3$: $3$, $6$, or $9$.
If we wrote in base $8$, instead of base $10$, then $7$ would have the property: if you write a number in base $8$ and add the digits (in base 8) until you get down to a single digit between $1$ and $7$, then the multiples of $7$ will always end yielding $7$, for precisely the same reason. And if we wrote in base $16$, then $15$ (or rather, F) would have the property. In general, if you write in base $b$, then $b-1$ has the property.
This is a special case of casting out nines, which in turn is a special case of modular arithmetic. It is what is behind many divisibility tests (e.g., for $2$, $3$, $5$, $9$, and $11$).
Coda. This reminds me of an anecdote a professor of mine used to relate: a student once came to him telling him he had discovered a very easy way to test divisibility of any number $N$ by any number $b$: write $N$ in base $b$, and see if the last digit is $0$. I guess, equivalently, you could write $N$ in base $b+1$, and add the digits to see if you get $b$ at the end.
An interesting case is $n=101$.
The $3$-digit palindrome $aba$ is divisible by 101 iff $b=0$.
The $4$-digit palindrome $abba$ is divisible by 101 iff $a=b$.
The $5$-digit palindrome $abcba$ is divisible by 101 iff $c=2a$.
The $6$-digit palindrome $abccba$ is divisible by 101 iff $a+b=c$.
The $7$-digit palindrome $abcdcba$ is divisible by 101 iff $d=2b$.
The $8$-digit palindrome $abcddcba$ is divisible by 101 iff $a+d=b+c$.
The $9$-digit palindrome $abcdedcba$ is divisible by 101 iff $e = 2(c-a)$.
The $10$-digit palindrome $abcdeedcba$ is divisible by 101 iff $a+b+e=c+d$.
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Best Answer
HINT: Suppose that you have a four-digit number $n$ that is written $abcd$. Then
$$\begin{align*} n&=10^3a+10^2b+10c+d\\ &=(999+1)a+(99+1)b+(9+1)c+d\\ &=(999a+99b+9c)+(a+b+c+d)\\ &=3(333a+33b+3c)+(a+b+c+d)\;, \end{align*}$$
so when you divide $n$ by $3$, you’ll get
$$333a+33b+3c+\frac{a+b+c+d}3\;.$$
The remainder is clearly going to come from the division $\frac{a+b+c+d}3$, since $333a+33b+3c$ is an integer.
Now generalize: make a similar argument for any number of digits, not just four. (If you know about congruences and modular arithmetic, you can do it very compactly.)