Solution:
The pairs sought are those for which one of the three conditions $a=b, a \nmid b$ or $b \nmid a$ is satisfied, i.e., all pairs of positive integers except those for which one of the numbers $a, b$ is a proper divisor of the other.
Throughout the entire proof we will denote the number of all divisors of a positive integer $m$ by $\tau(m)$. First it is clear that in the case $a=b$ one can choose an arbitrary $n$. Furthermore, in the case that $a$ is a proper divisor of $b$, for every positive integer $n$ the number $n a$ is a proper divisor of $n b$, so the set of its divisors is a proper subset of the set of divisors of $n b$, which implies $\tau(n a)<\tau(n b)$. The case $b \mid a, b<a$ is treated analogously.
It remains to show that in the case $a \nmid b, b \nmid a$ we can always find an $n$ with $\tau(n a)=\tau(n b)$. For this we first prove the following statement.
Lemma. Let $\alpha>\beta$ be positive integers. Then for every integer $M>\beta$ there exists a positive integer $\gamma$ such that
$$\frac{\alpha+\gamma+1}{\beta+\gamma+1}=1+\frac{1}{M}=\frac{M+1}{M} .$$
Proof. Rearranging the first equation gives $\gamma=M(\alpha-\beta)-(\beta+1) \geq 0$. Now we consider the prime factorization $a=\prod_{p} p^{\alpha_{p}}$ and $b=\prod_{p} p^{\beta_{p}}$ of $a$ and $b$. Let $p_{1}, \ldots p_{k}$ now be all those primes $p$ with $\alpha_{p}>\beta_{p}$ and $q_{1}, \ldots, q_{\ell}$ all primes $q$ with $\alpha_{q}<\beta_{q}$. Since $a \nmid b, b \nmid a$, both $k$ and $\ell$ are positive. Let $X$ now be a positive integer that is greater than all $\alpha_{p}$ and $\beta_{b}$. By the lemma above we can then choose numbers $\gamma_{1}, \ldots, \gamma_{k}$ and $\gamma_{1}^{\prime}, \ldots, \gamma_{\ell}^{\prime}$ such that
$$\frac{\alpha_{p_{i}}+\gamma_{i}+1}{\beta_{p_{i}}+\gamma_{i}+1}=\frac{k X+i}{k X+i-1}, \quad \text{ and } \quad \frac{\alpha_{q_{j}}+\gamma_{j}^{\prime}+1}{\beta_{q_{j}}+\gamma_{j}^{\prime}+1}=\frac{\ell X+j-1}{\ell X+j}$$
holds for all $1 \leq i \leq k$ and $1 \leq j \leq \ell$. For $n=\left(\prod_{i=1}^{k} p_{i}^{\gamma_{i}}\right) \cdot\left(\prod_{j=1}^{\ell} q_{j}^{\gamma_{j}^{\prime}}\right)$ it now follows, using the well-known formula
$$\tau\left(\prod_{p} p^{\eta_{p}}\right)=\prod_{p}\left(\eta_{p}+1\right)$$
that
$$\frac{\tau(n a)}{\tau(n b)}=\left(\prod_{i=1}^{k} \frac{\alpha_{p_{i}}+\gamma_{i}+1}{\beta_{p_{i}}+\gamma_{i}+1}\right) \cdot\left(\prod_{j=1}^{\ell} \frac{\alpha_{q_{j}}+\gamma_{j}^{\prime}+1}{\beta_{q_{j}}+\gamma_{j}^{\prime}+1}\right)=\frac{k(X+1)}{k X} \cdot \frac{\ell X}{\ell(X+1)}=1$$
so $n$ is as claimed.