The required integers are $n = 1$ and $n = 2$. In the former case, $(a_1, b_1) = (1, -2)$ is the unique pair of integers satisfying the conditions in the statement; and in the latter, only $(a_1, b_1, a_2, b_2) = (2, 0, -1, -2)$ and $(a_1, b_1, a_2, b_2) = (-1, -2, 2, 0)$ fit the bill. Verification is routine.

The degree $2n$ monic polynomial $f = \prod_{i=1}^{n} (X^2 + a_i X + b_i)$ vanishes at each $a_k$ and at each $b_k$. Since these $2n$ numbers are pairwise distinct and $\deg f = 2n$, they form the root set of $f$. Moreover, since the root set of each factor $f_i = X^2 + a_i X + b_i$ has size at most 2, and the union of these $n$ root sets has size $2n$, they must be pairwise disjoint sets of size 2 each. Recall now that $f$ is monic to write $f = \prod_{k=1}^{n} (X - a_k)(X - b_k)$. Consequently, $\prod_{k=1}^{n} (X^2 + a_k X + b_k) = \prod_{k=1}^{n} (X - a_k)(X - b_k)$.

If no $b_k$ is zero, identification of constant terms of both sides yields $a_1 \cdots a_n = 1$. This is impossible for pairwise distinct integers unless $n = 1$, in which case $a_1 = 1$, and $f = X^2 + X + b_1$. This polynomial must vanish at $a_1 = 1$, so $b_1 = -2$. Consequently, $f = X^2 + X - 2 = (X-1)(X+2) = (X-a_1)(X-b_1)$, as required.

If some $b_k = 0$, say, $b_1 = 0$, then $a_1 \neq 0$, so $f_1 = X^2 + a_1 X$ does not vanish at $a_1$. Since the root set of $f_1$ has size 2, this forces $n \ge 2$. The polynomial equality at the end of the second paragraph now reads $(X + a_1) \prod_{k=2}^{n} (X^2 + a_k X + b_k) = (X - a_1) \prod_{k=2}^{n} (X - a_k)(X - b_k)$.

Since $a_1, b_2, \dots, b_n$ are all non-zero, identification of constant terms of both sides above yields $a_2 \cdots a_n = -1$. This is impossible for pairwise distinct integers unless $n = 2$ or $n = 3$.

If $n = 2$, then $a_2 = -1$, and $f_1 = X^2 + a_1 X$ and $f_2 = X^2 - X + b_2$. Recall that the root sets of $f_1$ and $f_2$ are disjoint and both have size 2. Since $f_2(b_1) = f_2(0) = b_2 \neq b_1 = 0$ and $f_2(b_2) = b_2^2 \neq 0$, the roots of $f_2$ must be $a_1$ and $a_2$, so the other root of $f_1$ must be $b_2$. The condition $f_2(a_2) = 0$ yields $b_2 = -2$, and the condition $f_1(b_2) = 0$ then yields $a_1 = 2$. Consequently, $(a_1, b_1, a_2, b_2) = (2, 0, -1, -2)$.

The other quadruple, $(a_1, b_1, a_2, b_2) = (-1, -2, 2, 0)$, corresponds to the case where $b_2 = 0$.

Finally, to rule out the case $n = 3$, we may and will assume that $a_2 = -1$ and $a_3 = 1$. Then $f_1 = X^2 + a_1 X$, $f_2 = X^2 - X + b_2$, $f_3 = X^2 + X + b_3$, and $f = f_1 f_2 f_3$ has the (pairwise distinct) roots $a_1, a_2 = -1, a_3 = 1, b_1 = 0, b_2$ and $b_3$. Since $f_1(-1)f_1(1) = (1-a_1)(1+a_1) \neq 0$, it follows that $-1$ and $1$ are roots of $f_2 f_3$. Notice that $f(1) = b_2 \neq 0$ and $f_3(-1) = b_3 \neq 0$, so $2+b_2 = f_2(-1) = 0 = f_3(1) = 2+b_3$, i. e., $b_2 = b_3$. This contradiction settles the case.