14. Let $f(x)$ be an odd function defined on $\mathbf{R}$, and for any $x \in \mathbf{R}$, we have
$$
\begin{aligned}
f(x+2) & =f(x)+2, \\
\text { then } \sum_{k=1}^{2014} f(k) & =
\end{aligned}
$$
14.2029105.

Notice that, $f(0)=0$.
By $f(x+2)=f(x)+2$, let $x=-1$. Then $f(1)=f(-1)+2 \Rightarrow f(1)=1$.
$$
\begin{array}{l}
\text { Also, } f(2 n)=\sum_{k=1}^{n}(f(2 k)-f(2 k-2))+f(0) \\
=2 n, \\
f(2 n-1)=\sum_{k=2}^{n}(f(2 k-1)-f(2 k-3))+f(1) \\
=2 n-1, \\
\text { Therefore, } \sum_{k=1}^{2014} f(k)=\sum_{k=1}^{2014} k=2029105 .
\end{array}
$$