Let $ s(n)$ denote the sum of the digits of $ n$. Lemma. For positive integers $ a_1,a_2,...,a_n$ with sum $ S$, we have $ s(a_1)+s(a_2)+\cdots +s(a_n)\ge s(S)$ Proof. When summing the $ a_i$, if there is no "carrying" required, then the sum of the digits of $ S$ will be the same as the sum of the digits of all the $ a_i$. If any carrying is done, it will decrease the sum of the digits of $ S$ (by subtracting $ 10$ and adding $ 1$ to another digit), hence $ s(S)\le s(a_1)+s(a_2)+\cdots +s(a_n)$ as desired. Now, let $ a=10^n-1$, that is, the number with $ n$ digits, all nines. I claim that $ s(am)\ge 9n$ for any positive integer $ m$. I will prove this by strong induction on $ m$. The base case is $ m=1$. $ s(a)=9n$, as desired. Inductive step: Assume that this is true for all positive integers less than $ m$. Let: \[ am=d_0+d_1\cdot 10^n+d_2\cdot 10^{2n}+\cdots+d_k\cdot 10^{kn}\] with $ d_i<10^n$ for $ 0\le i\le k$. In other words, $ d_0,d_1,d_2,...$ are the digits in the base $ 10^n$ representation of $ am$. Since $ 10^n\equiv 1\mod a$, we have \[ d_0+d_1+\cdots+d_k\equiv d_0+d_1\cdot 10^n+d_2\cdot 10^{2n}+\cdots+d_k\cdot 10^{kn}\equiv am\equiv 0\mod a\] Thus there is some positive integer $ m'<m$ with $ am'=d_0+d_1+\cdots+d_k$, so by the inductive hypothesis $ s(d_0+d_1+\cdots+d_k)\ge 9n$. Finally, \[ s(am)=s(d_0)+s(d_1)+\cdots s(d_k)\ge s(d_0+d_1+\cdots+d_k)\ge 9n\] completing the inductive step. Thus, if we choose $ n>\frac{M}{9}$, then the arithmetic sequence $ a,2a,3a,...$ will have the sum of the digits of each term exceed $ M$.