Three, (50 points) Given the sequence $a_{1}, a_{2}, a_{3}, \cdots, a_{n}, \cdots, n \geqslant 1, a_{i} \in \mathbf{N}^{*}$, and $a_{a_{n}}+a_{n}=2 n$, find $a_{1}, a_{2}, a_{n}$.

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Three, Solution: For the condition $a_{a_{n}}+a_{n}=2 n$, taking $n=1$, we get $a_{a_{1}}+a_{1}=2$,
since $a_{1}, a_{a_{1}} \in \mathbf{N}^{*}$, so $a_{1}=1$, taking $n=2$, we get $a_{a_{2}}+a_{2}=4$,
since $a_{2}, a_{a_{2}} \in \mathbf{N}^{*}$, so $a_{2}=1,2,3$, if $a_{2}=1$, then $a_{1}+1=4$, we get $a_{1}=3$ which contradicts $a_{1}=1$, so it should be discarded; if $a_{2}=2$, then $a_{2}+2=4$, which is consistent; if $a_{2}=3$, then $a_{3}+3=4$, we get $a_{3}=1$,
then in the condition $a_{a_{n}}+a_{n}=2 n$ taking $n=3$, we get $a_{1}+1=6$, i.e., $a_{1}=5$ which contradicts $a_{1}=1$, so it should be discarded; in summary, we have: $a_{1}=1, a_{2}=2$.
Conjecture: $a_{n}=n$, and prove it by mathematical induction as follows:
(i) When $n=1$, the conjecture is obviously true;
(ii) Assume the conjecture holds for positive integers $n \leqslant k(k \geqslant 1)$, i.e., $a_{1}=1, a_{2}=2, \cdots, a_{k}=k$.
Now let $a_{k+1}=M \in \mathbf{N}^{*}$, and in the condition $a_{a_{n}}+a_{n}=2 n$ take $n=k+1$,
we get $a_{a_{k+1}}+a_{k+1}=2(k+1)$, i.e., $a_{M}+M=2 k+2$. (1)
If $M \leqslant k$, then by the induction hypothesis we know: $a_{M}=M$ substituting into (1) we get $M=k+1$, which is a contradiction, so it should be discarded;
If $M>k+1$, then from (1) we know: $a_{M}=2 k+2-M<2 M-(k+1)>k+1$. (4)
This (4) contradicts (3), so it should be discarded; thus, $M=k+1$, i.e., $a_{k+1}=k+1$, so when $n=k+1$ the conjecture also holds; combining (i) and (ii), we know, $a_{n}=n\left(n \in \mathbf{N}^{*}\right)$.