Let $a_1,a_2,\dots$ be an infinite sequence of positive integers. Suppose that there is an integer $N>1$ such that, for each $n\ge N$, the number

is an integer. Prove that there exists a positive integer $M$ such that $a_m=a_{m+1}$ for all $m\ge M$.

Step 1 (Setup): Let $S_n=a_1+a_2+\cdots +a_n$. We're given that $n|S_n$ for all $n\ge N$.

Step 2 (Key observation): Since $n|S_n$ and $(n+1)|S_{n+1}=S_n+a_{n+1}$, we get:
$(n+1)|(S_n+a_{n+1})$ and $n|S_n$.

Write $S_n=kn$ for some positive integer $k$. Then $S_{n+1}=kn+a_{n+1}$, and we need $(n+1)|(kn+a_{n+1})$.

Since $kn=k(n+1)-k$, this gives $(n+1)|(a_{n+1}-k)$, so $a_{n+1}\equiv k(\mathrm{mod}n+1)$.

Step 3 (Bounding $a_{n+1}$): For large $n$, $S_n/n\to$ some limit (since the average is always an integer and the $a_i$ are positive). Let $c=S_N/N$. As $n$ grows, the average $S_n/n$ cannot change too rapidly.

Since $a_{n+1}\equiv k_n(\mathrm{mod}n+1)$ where $k_n=S_n/n$, and $a_{n+1}$ is a positive integer, for large enough $n$ the only possibility is $a_{n+1}=k_n$ (since $a_{n+1}>0$ and the next valid value would be $k_n+(n+1)$, which would make the average jump).

Step 4 (Conclusion): Once $a_{n+1}=S_n/n$ for all large $n$, the average stays constant, forcing $a_{n+1}=S_n/n=S_{n-1}/(n-1)=\cdots$, which means all subsequent terms equal this constant. $■$