AustriaMO2013

We set $k:=\lfloor \frac{n}{a}\rfloor$ and $l:=\lfloor \frac{n+b}{a}\rfloor$. We thus have to find all nonnegative integers $n$ such that there exist integers $k$ and $l$ satisfying $$b+k=l\text{ and }k\le \frac{n}{a}<k+1\text{ and }l-1<\frac{n+b}{a}\le l.$$ By substituting $l=b+k$ we get the equivalent inequalities $$ka\le n<(k+1)a\text{ and }(b+k-1)a<n+b\le a(b+k).$$ Since all variables are integers we also have the equivalent relations $$ka\le n\le (k+1)a-1\text{ and }(b+k-1)a+1-b\le n\le a(b+k)-b.$$ Taking into account $$(b+k-1)a+1-b=ka+(a-1)(b-1)\ge ka$$ and $$a(b+k)-b=(k+1)a-1+(a-1)(b-1)\ge (k+1)a-1,$$ we see that the desired values of $n$ are exactly those satisfying $$(b+k-1)a+1-b\le n\le (k+1)a-1(1)$$ for some $k$. From (1) we conclude that for some $k$ we must have $(b+k-1)a+1-b\le (k+1)a-1$, that is $0\ge ab-2a+2-b=(a-1)(b-2)$. Thus it suffices to consider the following cases:

If $a=1$, (1) means $k\le n\le k$ and therefore all natural numbers $n\ge 0$ are solutions.

If $b=1$, (1) means $ka\le n\le (k+1)a-1$ and this is always satisfied for $k=\lfloor \frac{n}{a}\rfloor$. Hence all natural numbers $n\ge 0$ are solutions.

If $b=2$, (1) means $(k+1)a-1\le n\le (k+1)a-1$ and therefore necessarily $n=(k+1)a-1$. Such a $k$ can be found if and only if $n\equiv -1(\mathrm{mod}a)$, which are all the solutions in this case. (Note that for $a=1$ all integers $n\ge 0$ are solutions in accordance with the above.)

For $a\ge 2$ and $b\ge 3$ there is no solution.