Let's determine those natural numbers $a$ and $b$ for which

$$
90<a+b<100, \quad 0.9<\frac{a}{b}<0.91
$$
Multiplying the second inequality by $b$, then adding $b$ to it, we get $1.9 b < a + b < 1.91 b$. Combining this with the first inequality, we find that $90 < 1.91 b$ and $1.9 b < 100$. Therefore, $90 / 1.91 < b < 100 / 1.9$, and since $b$ is an integer, it follows that $48 \leq b \leq 52$.

We will check each of these five possible values to see if there is a corresponding $a$, that is, if there is an integer between $0.9 b$ and $0.91 b$.

If $b=48$, then $0.9 b=43.2$ and $0.91 b=43.68$, here there is no integer.

If $b=49$, then $0.9 b=44.1$ and $0.91 b=44.59$, here there is no integer.

If $b=50$, then $0.9 b=45$ and $0.91 b=45.5$, here there is no integer.

If $b=51$, then $0.9 b=45.9$ and $0.91 b=46.41$, here $a=46$.

If $b=52$, then $0.9 b=46.8$ and $0.91 b=47.32$, here $a=47$.

Since the obtained numbers satisfy the desired inequalities, the problem has two solutions:

$$
a=46, b=51, \quad \text{or} \quad a=47, b=52
$$