Solution:

There are $\binom{8}{4} \cdot 9^{4}$ sequences of 8 numbers with exactly four nines. A sequence of digits of length 8 is not an 8-digit number, however, if and only if the first digit is zero. There are $\binom{7}{4} 9^{3}$-digit sequences that are not 8-digit numbers. The answer is thus $\binom{8}{4} \cdot 9^{4} - \binom{7}{4} 9^{3} = 433755$.