Let $x_i=a_i+b_n$ then $\sum_{i=1}^n x_i=2n$ and $x_n \geq x_i \geq 1$ Note that $x_i^2+x_n^2\leq 1+(x_n+x_i-1)^2$ $n^2+3n=\sum_{k=1}^n (a_k^2+b_k^2)\leq \sum_{k=1}^n x_k^2 \leq 1*(n-1)+(x_1+...+x_n-(n-1))^2=n-1+(n+1)^2=n^2+3n$ so $x_1=x_2=..=x_{n-1}=1,x_{n}=n+1$ and so $(a_1,a_2,...,a_n),(b_1,b_2,...,b_n)=(0,0,...,0),(1,1,....,n+1)$

$x_i=a_i+b_i$ I guess @above

Let $a_k = 1 - \alpha_k$ and $b_k = 1 + \beta_k$,Then we have, $\sum\alpha_k = \sum\beta_k$ and $\sum\alpha_k^2 + \sum\beta_k^2 = n^2 + n$ Also, we have, for any $k$, $\alpha_k \le 1 \implies \sum\alpha_k \le n$ Also,$n + \sum\beta_k^2 \ge \sum\alpha_k^2 + \sum\beta_k^2 \ge n^2 + n$ $\implies \sum\beta_k^2 \ge n^2 \implies \sum\beta_k \ge n \implies \sum\alpha = n$ Hence, the only possible Solution is,$(a_1,a_2,...,a_n),(b_1,b_2,...,b_n)=(0,0,...,0),(1,1,....,n+1)$

One can see $b_n \leq n + 1$ easily. We have $(b_k - 1)(b_k - n - 1) \leq 0$, add for all $k$: \begin{align*} 0 &\geq \sum_{k = 1}^{n}(b_k - 1)(b_k - n - 1) \\ &= \sum_{k = 1}^{n}b_k^2 - (n+2)\sum_{k = 1}^{n}b_k + n^2 + n \\ &= n^2 + 3n - \sum_{k = 1}^{n}a_k^2 - (n + 2)\left(2n - \sum_{k = 1}^{n}a_k\right) + n^2 + n \\ &= (n + 2)\sum_{k = 1}^{n}a_k - \sum_{k = 1}^{n}a_k^2 \\ &\geq (n+2)\sum_{k = 1}^{n}a_k - \left(\sum_{k = 1}^{n}a_k\right)^2 \\ &= \left(\sum_{k = 1}^{n}a_k\right)\left(n + 2 - \sum_{k = 1}^{n}a_k\right) \end{align*} Note $\sum_{k = 1}^{n}a_k < n + 2$. We have $\sum_{k = 1}^{n}a_k = 0$ and equality holds. Therefore, either $b_k = 1$ or $b_k = n + 1$. Because of size issue, $b_1, b_2, \dots, b_{n - 1}$ must be $1$. Hence, the answer is $\boxed{(a_1, a_2, \dots, a_n) = (0, 0, \dots, 0), \ \ (b_1, b_2, \dots, b_n) = (1, 1, \dots, 1, n + 1)}$