Problem: Does there exist a positive integer $n$
such that $n$ has exactly 2000 distinct prime divisors
and $n$ divides $2^n + 1$?

Solution: Answer: Yes.

We say that $n$ is \emph{Korean} if $n \mid 2^n+1$.
First, observe that $n=9$ is Korean.
Now, the problem is solved upon the following claim:
\begin{claim*}
  If $n > 3$ is Korean,
  there exists a prime $p$ not dividing $n$
  such that $np$ is Korean too.
\end{claim*}
\begin{proof}
  I claim that one can take any primitive prime divisor $p$ of $2^{2n}-1$,
  which exists by Zsigmondy theorem.
  Obviously $p \neq 2$.
  Then:
  \begin{itemize}
    \ii Since $p \nmid 2^{\varphi(n)}-1$ it follows then that $p \nmid n$.
    \ii Moreover, $p \mid 2^n+1$ since $p \nmid 2^n-1$.
  \end{itemize}
  Hence $np \mid 2^n+1 \mid 2^{np} + 1$ by Chinese Theorem,
  since $\gcd(n,p) = 1$.
  % Thanks Wanlin for catching this
\end{proof}