AustriaMO

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
$$(b+k-1)a+1-b \le n \le (k+1)a-1 \quad (1)$$
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 $n \equiv -1 \pmod a$, 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.