4. Given the sequence $\left\{x_{n}\right\}$ satisfies $x_{1}=\frac{2}{3}, x_{n+1}=\frac{x_{n}}{2(2 n+1) x_{n}+1}(n=1,2, \cdots)$, then $\sum_{k=1}^{2016} x_{k}=$
4. $\frac{4032}{4033}$.

Clearly $x_{k} \neq 0$, by the problem we have
$$
\frac{1}{x_{k+1}}-\frac{1}{x_{k}}=4 k+2 \text {. }
$$

Therefore,
$$
\frac{1}{x_{n}}=\frac{1}{x_{1}}+\sum_{k=1}^{n-1}\left(\frac{1}{x_{k+1}}-\frac{1}{x_{k}}\right)=\frac{3}{2}+\sum_{k=1}^{n-1}(4 k+2)=\frac{4 n^{2}-1}{2} .
$$

Thus,
$$
\sum_{k=1}^{2016} x_{k}=\sum_{k=1}^{2016} \frac{2}{4 k^{2}-1}=\sum_{k=1}^{2016}\left(\frac{1}{2 k-1}-\frac{1}{2 k+1}\right)=1-\frac{1}{4033}=\frac{4032}{4033} .
$$