I know that the answer is N = 537, but I'm not sure how to solve for N analytically.
More precisely, my question is how can I simplify the sum of the binomial coefficients ?
The inequality is at this link:
https://i.stack.imgur.com/JE8HL.png
$$\left(\dfrac{1}{2}\right)^{1000} \sum_{x=N+1}^{1000} {1000 \choose x} < .01$$
You can get a good approximation using the normal approximation to the binomial distribution (with continuity correction): if $\Phi$ is the standard normal cumulative distribution function,
$$\left(\dfrac{1}{2}\right)^{1000} \sum_{x=N+1}^{1000} {1000 \choose x} \approx 1 - \Phi\left(\frac{N+1/2-500}{\sqrt{250}}\right)$$ Since $1 - \Phi(2.326347874) \approx 0.01$, this would lead to $N > 536.2827896$. So $N \ge 537$ is a pretty good guess. However, this is only approximate. Using exact rational arithmetic, the value for $N = 537$ does turn out to be slightly less than $.01$ (about $0.008831115668$) and for $N = 536$ slightly greater (about $0.01046355530$).
You can interpret this as saying that a certain window around the central binomial coefficient $\binom{1000}{500}$ has weight at least 98% of the total weight $2^{1000}$.
Since the binomial distribution tends to the normal distribution when suitably normalized, I expect you can get a good approximation for $N$ by looking at tables for the normal distribution: 98% occurs at the window of radius $2.33\sigma$ around the mean. Taking $\sigma=\sqrt{1000\cdot \dfrac12 \cdot \dfrac12}$, we get $N \approx \mu+2.33\sigma \approx 500+36.84 \approx 537$.
But I expect that an exact analytic solution is not known.
As hinted,
$$\left(\dfrac{1}{2}\right)^{1000} \sum_{x=N+1}^{1000} {1000 \choose x} < .01\\ \sum_{X=N+1}^{1000}{1000\choose X}\lt \frac 1{100}2^{1000}\\ \sum_{X=0}^{1000}{1000\choose X}\lt\frac 1{100}2^{1000}+\sum_{X=0}^{N}{1000\choose X}\\ 0.99\cdot 2^{1000}\lt \sum_{X=0}^{N}{1000\choose X}$$
Further analysis depends only on the RHS sum.
