Three, (20 points) Find all possible values of the positive integer $n$ such that for such $n$, there exist real numbers $a$ and $b$ for which the function $f(x)=\frac{1}{n} x^{2}+a x+b$ is an integer for any integer $x$.

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Three, let $f(x)=\frac{1}{n} x^{2}+a x+b$ be such that for any integer $x, f(x)$ is an integer. Then,
$$
\begin{array}{l}
g(x)=f(x+1)-f(x) \\
=\left[\frac{1}{n}(x+1)^{2}+a(x+1)+b\right]- \\
\quad\left(\frac{1}{n} x^{2}+a x+b\right) \\
=\frac{2}{n} x+\frac{1}{n}+a
\end{array}
$$

is also an integer. Furthermore,
$$
g(x+1)-g(x)=\frac{2}{n}
$$

is also an integer.
Therefore, $n=1$ or 2.
When $n=1$, take integers $a, b$, then $f(x)=x^{2}+$ $a x+b$ is an integer for any integer $x$.
When $n=2$, take $a=\frac{1}{2}, b$ as an integer, then
$$
f(x)=\frac{1}{2} x^{2}+\frac{1}{2} x+b=\frac{1}{2} x(x+1)+b
$$

is an integer for any integer $x$.
In summary, $n=1$ or 2.