A special year for me, back in time!

To have the smallest $n$ , it is obvious that we pack as many 9's as possible as its digits so that $n$ is "shortest" with the least number of digits. The maximum number of 9's possible is given by $\left \lfloor \dfrac {2012}9 \right \rfloor = 223$ . The remaining digit is $2012-223\times 9 = 5$ . Therefore, $n=5\underbrace{99999\cdots 99999}_{\text{Number of 9's}= 223}$ and $n+1 = 6\underbrace{00000\cdots00000}_{\text{Number of 0's}= 223}$ . The first digit of $n+1$ is $\boxed 6$ .

To achieve the lowest possible number, we first need to minimize the amount of digits, and then make the first digit as small as possible. The way to do that would be to make all digits except the first digit 9. The first number lower than 2012 that is divisible by 9 is 2007. 2012 divided by 2007 leaves a remainder of 5, so the first digit is 5, followed by a number of 9’s. When we add 1, it carries all the way to the first digit, because all the digits after it are 9’s. Therefore we need to add 1 to 5, to achieve an answer of $\boxed{6}$ .