Three, (50 points) Find the largest positive integer $n$, such that for any integer $a$, if $(a, n)=1$, then $a^{2} \equiv 1(\bmod n)$.
Three, the maximum value of $n$ is 24.
First, verify that $n=24$ holds.
For any $(a, 24)=1$, we have
$(a, 3)=1 \Rightarrow a^{2} \equiv 1(\bmod 3)$,
and $(a, 2)=1 \Rightarrow a^{2} \equiv 1(\bmod 8)$,
thus, $a^{2} \equiv 1(\bmod 24)$.
Next, prove that $n \leqslant 24$. Consider two cases.
(1) If 5 † $n$, then take $a=5$.
Since $a^{2} \equiv 1(\bmod n)$, we have $25 \equiv 1(\bmod n)$.
Therefore, $n \leqslant 24$.
(2) If $5 \mid n$, assume without loss of generality that $n=5^{\alpha} 2^{\beta} \gamma(\alpha \geqslant 1, \beta \geqslant 0, \gamma$ is coprime with 10).
Take $a=5 \gamma+2$. Then
$$
\begin{array}{l}
(a, \gamma)=(2, \gamma)=1, \\
(a, 5)=1,(a, 2)=1 .
\end{array}
$$

Thus, $(a, n)=1$, and $a^{2} \equiv 4(\bmod 5)$, which contradicts $a^{2} \equiv 1(\bmod n)$.