The answer is 333667.

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First, we realized that digit sum of unities,tens,hundreds...columns it's the same. There are $9!$ numbers of this kind, so each column had $8!$ times the numbers $1$ to $9$ .

Sum of each colum is $\frac{(9)(10)}{2}(8!)(10^{k-1})$ and $k$ is the column number.

The sum of all columns is $N$ and $N=\frac{(9)(10)}{2}(8!)(10^{0})$ + $\frac{(9)(10)}{2}(8!)(10^{1})$ + $\frac{(9)(10)}{2}(8!)(10^{3})$ +...+ $\frac{(9)(10)}{2}(8!)(10^{8})$ .

Factoring $N=8!\times 45\times111,111,111$

Using divisibility for $111,111,111$ , $111,111,111=9\times 37\times 333667$

$N=2^7×3^6×5^2×7×37×333667 \therefore \boxed{333667}$ is the greatest prime factor of $N$ .