1. 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 $x_{1}+x_{2}+\cdots+x_{2016}$.
From condition (1), we get: $\left|x_{k}\right| \leqslant 1,\left|y_{k}\right| \leqslant 1, k=1,2, \cdots, 2016$, thus (note $x_{k} \geqslant 0$)
$$
\sum_{k=1}^{2016} x_{k} \geqslant \sum_{k=1}^{2016} x_{k}^{2}=\sum_{k=1}^{2016}\left(1-y_{k}^{2}\right)=2016-\sum_{k=1}^{2016} y_{k}^{2} \geqslant 2016-\sum_{k=1}^{2016}\left|y_{k}\right| \text {. }
$$

Without loss of generality, assume $y_{1}, y_{2}, \cdots, y_{m}>0, y_{m+1}, \cdots, y_{2016} \leqslant 0,0 \leqslant m \leqslant 2016$, then
$$
\begin{array}{l}
\sum_{k=1}^{2016}\left|y_{k}\right|=\sum_{k=1}^{m} y_{k}-\sum_{k=m+1}^{2016} y_{k} \leqslant 2016 \Rightarrow \sum_{k=1}^{m} y_{k} \leqslant m,-\sum_{k=m+1}^{2016} y_{k} \leqslant 2016-m . \\
\text { If } \sum_{k=1}^{m} y_{k}>m-1 \text {, and }-\sum_{k=m+1}^{2016} y_{k}>2015-m \text {, let } \\
\sum_{k=1}^{m} y_{k}=m-1+a,-\sum_{k=m+1}^{2016} y_{k}=2015-m+b \text {, then } \\
\left\{\begin{array}{l}
m-1<m-1+a \leqslant m, \\
2015-m<2015-m+b \leqslant 2016-m
\end{array} \Rightarrow 0<a, b<1\right. \text {, thus } \\
\sum_{k=1}^{2016} y_{k}=\sum_{k=1}^{m} y_{k}+\sum_{k=m+1}^{2016} y_{k}=(m-1+a)-(2015-m+b)=2 m-2016+a-b,
\end{array}
$$

From condition (2), we know $\sum_{k=1}^{2016} y_{k}$ is an odd number, thus $a-b$ is an odd number, which contradicts $0<a, b<1$.
Therefore, there must be $\sum_{k=1}^{m} y_{k} \leqslant m-1$ or $-\sum_{k=m+1}^{2016} y_{k} \leqslant 2015-m$, then
$\sum_{k=1}^{2016}\left|y_{k}\right|=\sum_{k=1}^{m} y_{k}-\sum_{k=m+1}^{2016} y_{k} \leqslant m-1+2016-m$ or $m+2015-m=2015$.
Thus, combining (1) we get $\sum_{k=1}^{2016} x_{k} \geqslant 1$. Also, when $x_{1}=x_{2}=\cdots=x_{2015}=0, x_{2016}=1$;
$y_{1}=y_{2}=\cdots=y_{2015}=1, y_{2016}=0$ satisfies the conditions, and makes the equality in the above inequality hold, so the minimum value of $x_{1}+x_{2}+\cdots+x_{2016}$ is 1.