Two rational numbers, as well as the sum of their reciprocals, are integers. What could the numbers be?
We claim that if the sum of two rational numbers is an integer, then in their simplest form, the absolute values of the denominators are equal.

Indeed, if $\left(p_{1}, q_{1}\right)=1,\left(p_{2}, q_{2}\right)=1$ and $\frac{p_{1}}{q_{1}}+\frac{p_{2}}{q_{2}}$ is an integer, then

$$
q_{1} q_{2} \mid\left(p_{1} \cdot q_{2}+p_{2} \cdot q_{1}\right)
$$

From this, on one hand, $q_{1} \mid p_{1} \cdot q_{2}$, and since $\left(p_{1}, q_{1}\right)=1$, it follows that $q_{1} \mid q_{2}$. On the other hand, similarly, $q_{2} \mid q_{1}$. Therefore, one denominator is a divisor of the other, and their absolute values are indeed equal.

According to the condition, the sum of the reciprocals of the numbers is also an integer, so in their simplest form, the absolute values of the numerators are also equal. We conclude that the absolute values of the two rational numbers are equal.

Clearly, any non-zero rational number and its negative satisfy this, since in this case both sums are 0.

If the numbers are equal, denoting their common value by $r$, the condition implies that $2 r$ and $2 / r$ are both integers. Since $2 / r=4 / 2 r$, we get that $2 r$ is a divisor of 4. Hence, $|r|=2$, or $|r|=1$, or $|r|=\frac{1}{2}$; and it can be seen that in each case the condition is indeed satisfied.

Therefore, the pairs of numbers we are looking for, besides the pairs of the form $(x,-x)$ where $x$ is a non-zero rational number, are $(2,2) ;(-2,-2) ;(1,1) ;(-1,-1) ;\left(\frac{1}{2}, \frac{1}{2}\right)$ and $\left(-\frac{1}{2},-\frac{1}{2}\right)$.