Using familiar lemma

Let $\displaystyle l_m=\underset{k\ge m}{\mathop{\inf }}\pi \left( k \right)\frac{\ln k}{k}, u_m=\underset{k\ge m}{\mathop{\sup }}\pi \left( k \right)\frac{\ln k}{k}$ then $l_m\to1$ and $u_m\to 1$ by the Prime Number Theorem.

For $n\ge k\ge m >1$ , using $\dfrac{k}{n}\ln\dfrac{n}{k}\le\dfrac{1}{e}$ we have

$\begin{array}{lll} &n+\pi\left(k\right)\ln n &\ge &n+l_mk\dfrac{\ln n}{\ln k}\ge n+l_mk \\ &n+\pi\left(k\right)\ln n &\le &n+u_mk+u_mk\dfrac{\ln \dfrac{n}{k}}{\ln k}\le n+u_mk+u_m\dfrac{n}{e\ln m} \end{array}$

Then

$\begin{array}{rl} \displaystyle\underset{n\to \infty }{\mathop{\lim \sup }}\sum\limits_{k=1}^{n}{\frac{1}{n+\pi \left( k \right)\ln n}}&\displaystyle=\underset{n\to \infty }{\mathop{\lim \sup }}\sum\limits_{k=m}^{n}{\frac{1}{n+\pi \left( k \right)\ln n}} \\ &\displaystyle \le \underset{n\to \infty }{\mathop{\lim \sup }}\sum\limits_{k=m}^{n}{\frac{1}{n+l_m k}} \\ &\displaystyle =\underset{n\to \infty }{\mathop{\lim \sup }}\frac{1}{n}\sum\limits_{k=1}^{n}{\frac{1}{1+l_m\frac{k}{n}}}=\int\limits_0^1 \frac{\text{d}x}{1+l_m x} \end{array}$

In a similar way we get $\displaystyle\underset{n\to \infty }{\mathop{\lim \inf }}\sum\limits_{k=1}^{n}{\frac{1}{n+\pi \left( k \right)\ln n}}\ge \int\limits_0^1\frac{\text{d}x}{1+u_m x+\dfrac{u_m}{e\ln m}}$

Taking $m\to\infty$ , the limit is $\displaystyle\int_0^1 \frac{\text{d}x}{1+x}=\ln 2$ .

So $\left\lceil 1000L\right\rceil =\boxed{694}$ .