If $f = (X - x_{1})(X - x_{2}),\ x_{1}, x_{2} \in \mathbb{Z}$, then take $u_{n} = n + x_{1}$ and get $f\left(u_{n}\right) = n\left(n + x_{1} - x_{2}\right)$, hence $n \mid f\left(u_{n}\right),\ n \geq 1$.

Conversely, assume that $f\left(u_{n}\right) = k_{n} \cdot n$, for some integer $k_{n}$, $n = 1, 2, \ldots$ Then, the quadratic equation
$$u^{2} + a \nu + b - k_{n} \cdot n = 0$$
has integer zeros, hence its discriminant $\Delta_{n}$ is a perfect square, that is $\Delta_{n} = t_{n}^{2},\ n = 1, 2, \ldots$. This is equivalent to
$$a^{2} - 4\left(b - k_{n} \cdot n\right) = t_{n}^{2}$$
hence we have
$$\Delta + 4 k_{n} \cdot n = t_{n}^{2}, \quad n = 1, 2, \ldots \tag{1}$$
where $\Delta = a^{2} - 4b$ is the discriminant of equation $f(n) = 0$. In (1) we take $n = \Delta^{2}$ and obtain
$$\Delta \cdot \left(1 + 4 k_{\Delta^{2}} \cdot \Delta\right) = t_{\Delta_{2}}^{2} . \tag{2}$$
Since $\Delta$ and $1 + 4 k_{\Delta^{2}} \cdot \Delta$ are relatively prime, from (2) it follows that $\Delta$ is a perfect square, hence the zeros of $f$, $x_{1}, x_{2}$ are rational. We have $\Delta = a^{2} - 4b = c^{2},\ c \in \mathbb{Z}$, and
$$x_{1,2} = \frac{1}{2}(-a \pm c) . \tag{3}$$
It is clear that $a$ and $c$ are of the same parity, and from (3) we get that $x_{1}, x_{2} \in \mathbb{Z}$.