Euler, is that you?

Let $X$ be a discrete random variable that follows the Poisson distribution with expected value $n.$ We have

$\begin{aligned} P(X \leq n) &= \sum_{i = 0}^n P(X = i) \\ &= \sum_{i = 0}^n \dfrac{e^{-n}n^i}{i!} \\ &= e^{-n} \sum_{i = 0}^n \dfrac{n^i}{i!}. \end{aligned}$

Thus, $N = \lim_{n \rightarrow \infty} P(X \leq n).$ As $n$ gets larger and larger, $X$ will approach a normal distribution, so $\lim_{n \rightarrow \infty} P(X \leq n)$ will just be equal to the fraction of the total area under a normal curve that is to the left of the mean. This is clearly equal to $\dfrac{1}{2},$ therefore $N = \dfrac{1}{2}$ and $\lfloor 1000N \rfloor = \left \lfloor 1000 \left ( \dfrac{1}{2} \right ) \right \rfloor = \boxed{500}.$