4. The sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=2, a_{n+1}=\frac{2(n+2)}{n+1} a_{n}\left(n \in \mathbf{N}^{*}\right)$, then $\frac{a_{2014}}{a_{1}+a_{2}+\cdots+a_{2013}}$
$$
\begin{array}{l}
\frac{a_{n+1}}{a_{n}}=\frac{2(n+2)}{n+1} \Rightarrow a_{n}=\frac{a_{n}}{a_{n-1}} \cdot \frac{a_{n-1}}{a_{n-2}} \cdots \frac{a_{3}}{a_{2}} \cdot \frac{a_{2}}{a_{1}} \cdot a_{1}=\frac{2^{n-1}(n+1) n \cdots 4 \cdot 3}{n(n-1) \cdots 3 \cdot 2} \cdot 2 \\
=(n+1) \cdot 2^{n-1}\left(n \in \mathbf{N}^{*}\right)
\end{array}
$$

Let $S=a_{1}+a_{2}+\cdots+a_{n}=2+3 \cdot 2+4 \cdot 2^{2}+\cdots+(n+1) \cdot 2^{n-1}$, then $2 S=2 \cdot 2+3 \cdot 2^{2}+\cdots+n \cdot 2^{n-1}+(n+1) \cdot 2^{n}$, subtracting the two equations gives
$$
\begin{array}{l}
-S=2+2+2^{2}+\cdots+2^{n-1}-(n+1) \cdot 2^{n}=2^{n}-(n+1) \cdot 2^{n}=-n \cdot 2^{n} \\
\Rightarrow S=n \cdot 2^{n} \Rightarrow a_{1}+a_{2}+\cdots+a_{2013}=2013 \cdot 2^{2013} .
\end{array}
$$

Therefore, $\frac{a_{2014}}{a_{1}+a_{2}+\cdots+a_{2013}}=\frac{2015 \cdot 2^{2013}}{2013 \cdot 2^{2013}}=\frac{2015}{2013}$.