Three, (50 points) Find the smallest positive integer $n$, such that the sum of the squares of all positive divisors of $n$ is $(n+3)^{2}$.

 untranslated part:

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Note: The last sentence is a note and not part of the problem statement, so it is provided in its original form as it is not to be translated.
Three, let $16 n+8, \\
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Contradiction.
Thus, $k \leqslant 5$.
And $k=0$ clearly does not meet the problem's conditions, so $1 \leqslant k \leqslant 5$.
Let $n=\prod_{i=1}^{i} p_{i}^{\alpha_{i}}$. Then
$$
\prod_{i=1}^{t}\left(\alpha_{i}+1\right)=k+2 \in[3,7] .
$$

Thus, $1 \leqslant t \leqslant 2$.
We discuss three cases.
(1) $n=p^{\alpha}$ (where $p$ is a prime, $\alpha \in \mathbf{Z}_{+}$, and $2 \leqslant \alpha \leqslant 6$).

Then $p^{2}+p^{4}+\cdots+p^{2(\alpha-1)}=6 p^{\alpha}+8$.
Thus, $p^{2} 18 \Rightarrow p=2$.
Taking both sides modulo 8, we get $4 \equiv 0(\bmod 8)$, contradiction.
(2) $n=p q$ (where $p<q$ are primes).

Then $p^{2}+q^{2}=6 p q+8$.
Thus, $q=3 p+2 \sqrt{2 p^{2}+2}$.
Upon verification, the smallest prime $p=7$ makes $q$ a prime, at which point, $q=41, n=p q=287$.
(3) $n=p q^{2}$ (where $p, q$ are distinct primes).
Then $p^{2}+q^{2}+q^{4}+p^{2} q^{2}=6 p q^{2}+8$.
If $2 \mid n$, taking both sides of equation (1) modulo 4, we find $p=q=2$, contradiction.

Otherwise, $p, q$ are both odd, taking both sides of equation (1) modulo 8, we get $4 \equiv 6 p(\bmod 8)$, so $p=2$, contradiction.
In summary, the minimum value of $n$ is 287.