Mistranslating! The problem should be $1+C_n^1+C_n^2+C_n^3\mid 2^{2000}$.

agzqu wrote: Mistranslating! The problem should be $1+C_n^1+C_n^2+C_n^3\mid 2^{2000}$. I have corrected the mistake. Thank you!

We want $1+C_n^1+C_n^2+C_n^3 = \dfrac {1} {6}(n+1)(n^2-n+6) \mid 2^{2000}$. So $n=2^k-1$ or $n=3\cdot 2^k-1$. Now it's easy to see the only solutions are $n=2^3-1=7$ and $n=3\cdot 2^3-1 = 23$ (for larger values LHS has an odd divisor).

This is an obvious reference of the https://en.wikipedia.org/wiki/Binary_Golay_code .