Lemma. If $a\mid 2^a+1$ and $b\mid 2^b+1$ , where $a=3^\alpha a'$ , $b=3^\beta b'$ , $3 \nmid a', b'$ and $a'$ and $b'$ are coprime, then $ab\mid 2^{ab}+1$ .
Proof. Obviously, $a$ and $b$ are odd.
$2^{ab}\equiv (-1)^b\equiv -1 \pmod{a}$ $2^{ab}\equiv (-1)^a\equiv -1 \pmod{b}$ By LTE, $v_3(2^{ab}+1)=v_3(ab)+1=\alpha + \beta + 1$ . Finally, as $a', b', 3^{\alpha+\beta}\mid 2^{ab}+1$ and all three numbers are pairwise coprime, $ab=a'b'3^{\alpha+\beta}\mid 2^{ab}+1$ . $\square$

We construct a sequence of primes $p_2, p_3, \dots, p_{2000}$ in the following manner. Let $p_2=19$ (so that $p_2\mid 2^{3^2}+1$ ). Inductively assume we have constructed $p_2, p_3, \dots, p_{i}$ , for some $i\ge 2$ , such that:
$\bullet$ all $p_j$ with $2\le j\le i$ are distinct,
$\bullet$ $p_j>3$ and $p_j\mid2^{3^{i}}+1$ for each $2\le j\le i$ .
As $2^{3^{i+1}}+1=(2^{3^i}+1)(4^{3^i}-2^{3^i}+1)$ and the greatest common divisor of the terms in the parenthesis is $3$ , we conclude that $4^{3^i}-2^{3^i}+1$ has all its prime divisors distinct from $p_j$ for $2\le j\le i$ . As $9\mid 2^{3^i}+1$ and $4^{3^i}-2^{3^i}+1>3$ , it has a prime factor $p_{i+1}>3$ , so the inductive claim holds for $i+1$ as well.

Having constructed $p_2, p_3, \dots, p_{2000}$ , let $n_i=3^i p_i$ for each $2\le i\le 2000$ . By LTE, $v_3(2^{n_i}+1)=i+1$ . By Fermat's Little Theorem, $2^{n_i}\equiv 2^{3^i p_i}\equiv 2^{3^i}\equiv -1\pmod{p_i}$ . Therefore, $n_i = 3^i p_i\mid 2^{n_i} + 1$ for each $i$ . Let $n=n_2n_3\dots n_{2000}$ . By repeated application of the lemma, $n\mid 2^n + 1$ , and by construction, $n$ has $2000$ distinct prime divisors.