Solution:

Each summand is $10^k - 1,$ so $$N = \sum_{k=1}^{321} \left(10^k - 1\right) = \underbrace{11\ldots1}_{321}0 - 321.$$

The subtraction changes only the last four digits: $1110 - 321 = 789,$ so those four digits become $0789.$ Thus $N$ consists of $318$ ones followed by $0789,$ and the digit sum is $318 + 0 + 7 + 8 + 9 = 342.$