One, (This question is worth 40 points) Non-negative real numbers $x_{1}, x_{2}, \cdots, x_{2016}$ and real numbers $y_{1}, y_{2}, \cdots, y_{2016}$ satisfy:
(1) $x_{k}^{2}+y_{k}^{2}=1, k=1,2, \cdots, 2016$;
(2) $y_{1}+y_{2}+\cdots+y_{2016}$ is an odd number.

Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2016}$.
Solution: From the given condition (1), we have: $\left|x_{k}\right| \leq 1,\left|y_{k}\right| \leq 1, k=1,2, \cdots, 2016$, thus (note $x_{i} \geq 0$)
$$
\sum_{k=1}^{2016} x_{k} \geq \sum_{k=1}^{2016} x_{k}^{2}=\sum_{k=1}^{2016}\left(1-y_{k}^{2}\right)=2016-\sum_{k=1}^{2016} y_{k}^{2} \geq 2016-\sum_{k=1}^{2016}\left|y_{k}\right| \text {. }
$$
10 points

Without loss of generality, assume $y_{1}, \cdots, y_{m}>0, y_{m+1}, \cdots, y_{2016} \leq 0,0 \leq m \leq 2016$. Then
$$
\begin{array}{c}
\sum_{k=1}^{m} y_{k} \leq m,-\sum_{k=m+1}^{2016} y_{k} \leq 2016-m . \\
\text { If } \sum_{k=1}^{m} y_{k}>m-1, \text { and }-\sum_{k=m+1}^{2016} y_{k}>2015-m, \text { let } \\
\sum_{k=1}^{m} y_{k}=m-1+a,-\sum_{k=m+1}^{2016} y_{k}=2015-m+b,
\end{array}
$$

then $0<a, b<1$, thus
$$
\begin{aligned}
\sum_{k=1}^{2016} y_{k} & =\sum_{k=1}^{m} y_{k}+\sum_{k=m+1}^{2016} y_{k}=m-1+a-(2015-m+b) \\
& =2 m-2016+a-b,
\end{aligned}
$$

From condition (2), we know that $\sum_{k=1}^{2016} y_{k}$ is an odd number, so $a-b$ is an odd number, which contradicts $0<a, b<1$.
Therefore, there must be $\sum_{k=1}^{m} y_{k} \leq m-1$, or $-\sum_{k=m+1}^{2016} y_{k} \leq 2015-m$, then
$$
\sum_{k=1}^{2016}\left|y_{k}\right|=\sum_{k=1}^{m} y_{k}-\sum_{k=m+1}^{2016} y_{k} \leq 2015 \text {. }
$$

Thus, combining (1) we get $\sum_{k=1}^{2016} x_{k} \geq 1$.
Furthermore, when $x_{1}=x_{2}=\cdots=x_{2015}=0, x_{2016}=1, y_{1}=y_{2}=\cdots y_{2015}=1, y_{2016}=0$ the conditions of the problem are satisfied, and the equality in the inequality holds, so the minimum value of $x_{1}+x_{2}+\cdots+x_{2016}$ is 1.
40 points