How about induction?

It is obviously true for the one-digit numbers $3, 6$ and $9$, so we have our base case (really, just the case $3$ is all it takes, but I like to be on the safe side when it comes to induction).

Now, let's say that we have a number divisible by $3$, and let's call it $n$. We can also assume that the sum of the digits of $n$ is divisible by $3$. I want to show that the sum of digits of $n+3$ is also divisible by $3$. If that is the case, then we are done, for the induction principle takes care of any case for us from there.

The sum of the digits of $n$ is some number, let's call it $m$, and this number is assumed to be divisible by $3$. Now, if we're lucky, the sum of digits in $n+3$ is just $m+3$, and by lucky I mean there is no carry involved. So, if there is no carry involved in adding $3$ to $n$, then we are done.

If there is a carry, however, then let's pretend for a second that the last digit of $n$ can surpass $9$. Were that the case, the sum of digits of $n+3$ would really *be* $m+3$. This is sadly not the case, but what really happens when we do the carry? We subtract $10$ from the $1$-digit, and add $1$ to the $10$-digit. This will have the net effect on the sum of digits that we subtract $9$, so in that case the sum of digits in $n+3$ is $m+3-9 = m-6$, which is still divisible by $3$, so there is no problem!

"Hold on there, not so fast", you say. "What if adding $1$ to the $10$-s digit makes a carry happen there?" Well, my enlightened reader, in that case the same argument as in the paragraph above would apply, only moved one space to the left in the digits of $n$. The net effect: the sum of digits of $n+3$ is $m-6-9 = m-15$, still divisible by $3$. If there is a carry from the hundreds-digit, then we will subtract another $9$ for a total of $m-24$. And so on. You will never make a carry like that take $m+3$ out of divisible-by-three-space. And this concludes the proof.