Splendid problem. We have $b+k\mid a-b$ for $1\le k\le b-1$, and we must prove $k\mid a-b$ for $1\le k\le b-1$. For this, it suffices to show that for each $k\le b-1$, there is an $i$ such that $b+1\le ki\le 2b-1$. Assume this is false for some $k$. As $ki=k\le b-1$ for $i=1$, it is the case that there is an $i_0$ such that $ki_0\le b$ and $k(i_0+1)\ge 2b$. But then $k\ge 2b-b=b$, a contradiction.

$b+k|a+k \to b+k| a-b$ for all $k<b$ Let $n$ is such natural number that $\frac{k}{b-k}<n<\frac{b+k}{b-k}$( such number exists because $\frac{b+k}{b-k}-\frac{k}{b-k}>1)$ Let $m=n(b-k)-k$ then $0<m<b$ $b+m=b+n(b-k)-k=(n+1)(b-k)$ and so $b-k|b+m \to b-k|a-b \to b-k|a-k$