$\Theta \rho ϵ\eta \Pi \alpha \tau \pi$

First note that it's impossible for $n>\frac{M}{ϵ}$, since if it were true we would know that: $$\sum _{i=0}^n{a}_i\ge nϵ>\frac{M}{ϵ}ϵ=M$$ but we require this sum to be equal to $M$, therefore we know that $n\le \frac{M}{ϵ}$. So we have an upper bound on the value of $n$

Suppose that $ϵ|M$, therefore $\frac{M}{ϵ}\in \mathbb{Z}$, and thus by taking $n=\frac{M}{ϵ}$ then clearly $n$ is at a maximum as needed, and by taking $a_i=ϵ$ we have $$\sum _{i=1}^n{a}_i=nϵ=M$$ Also note that $\lfloor \frac{M}{ϵ}\rfloor =\frac{M}{ϵ}$ in this case.

Otherwise $ϵ\nmid M$, therefore $\frac{M}{ϵ}\in ℝ\setminus \mathbb{Z}$, therefore we can certainly not pick $n=\frac{M}{ϵ}$, and as stated earlier we cannot have $n$ exceed this value, so the next largest value would the largest integer less than or equal to $\frac{M}{ϵ}$ which is defined as $\lfloor \frac{M}{ϵ}\rfloor$

Recall that by the quotient remainder theorem there exists some $q,r\in \mathbb{Z}$ such that $q\cdot ϵ+r=M$, and that specifically we have that $q=\lfloor \frac{M}{ϵ}\rfloor$ and $r=M-\lfloor \frac{M}{ϵ}\rfloor ϵ$ which yields the equation

$$\lfloor \frac{M}{ϵ}\rfloor \cdot ϵ+(M-\lfloor \frac{M}{ϵ}\rfloor ϵ)=(\lfloor \frac{M}{ϵ}\rfloor -1)\cdot ϵ+(M-\lfloor \frac{M}{ϵ}\rfloor ϵ+ϵ)$$ And note that since $r\ge 0$, well that is $M-\lfloor \frac{M}{ϵ}\rfloor \ge 0$, we know that $$M-\lfloor \frac{M}{ϵ}\rfloor ϵ+ϵ>ϵ$$ and thus the sequence given by $a_i=ϵ$ for $i\in [1\dots (\lfloor \frac{M}{ϵ}\rfloor -1)]$ and $$a_{\lfloor \frac{M}{ϵ}\rfloor }=(M-\lfloor \frac{M}{ϵ}\rfloor ϵ+ϵ)$$ is clearly bounded below by $ϵ$, and has sum equal to $M$, and has $\lfloor \frac{M}{ϵ}\rfloor$ terms, so the largest value of $n$ is preciely that.