Answer. For $n=6$.

Solution. For $n=6$, we can set $a_{1}=a_{2}=a_{3}=a_{4}=1$ and $a_{5}=a_{6}=-1$; then the quadratic trinomial from the condition becomes $x^{2}-8 x+7$ and has two integer roots: 1 and 7. It remains to show that this is the smallest possible value of $n$.

Suppose the numbers $a_{1}, a_{2}, \ldots, a_{n}$ satisfy the condition of the problem; then the discriminant of the quadratic trinomial from the condition, divided by 4, must be a perfect square. It is equal to

$$
d=\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{4}-\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$

Then the number $d$ is odd and is a square, so it gives a remainder of 1 when divided by 8.

Rewrite the above equality as

$$
d+1+a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}=\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{4}
$$

and consider it modulo 8. It is not hard to check that the fourth powers of integers give only remainders 0 and 1 when divided by 8, so the right-hand side of the equality gives a remainder of 0 or 1. The left-hand side, however, is comparable to $1+1+k$, where $k$ is the number of odd numbers among $a_{i}$. Therefore, $n \geqslant k \geqslant 6$.
9.2. For what least natural $n$ do there exist integers $a_{1}, a_{2}, \ldots, a_{n}$ such that the quadratic trinomial

$$
x^{2}-2\left(a_{1}+a_{2}+\ldots+a_{n}\right)^{2} x+\left(a_{1}^{4}+a_{2}^{4}+\ldots+a_{n}^{4}+1\right)
$$

has at least one integer root?

(P. Kozlov)