Let $ k = n+2005 $. Since $ 1 \le n \le 9999 $, we have $ 2006 \le k \le 12004 $. We know that $ k $ has exactly 21 positive factors. The number of positive factors of a positive integer with prime factorization $ p_1^{e_1}p_2^{e_2} \cdots p_r^{e_r} $ is $ (e_1+1)(e_2+1)\cdots(e_r+1) $. Since $ 21 = 7 \cdot 3 $ and 7 and 3 are prime, the prime factorization of $ k $ is either of the form $ p^{20} $ or $ p^6 q^2 $, where $ p $ and $ q $ are distinct prime numbers. Since $ p^{20} \geq 2^{20} > 12004 $ for any prime $ p $, we can't have the first form. So $ k = p^6 q^2 $ for distinct primes $ p $ and $ q $. If $ p=2 $, then $ k=64q^2 $. So $ 2006 \le 64q^2 \le 12004 \Rightarrow 31.34375 \le q^2 \le 187.5625 $. For $ q $ an integer, this holds when $ 6 \le q \le 13 $. Since $ q $ is prime, $ q $ is 7, 11, or 13. So if $ p=2 $, the possible values of $ k $ are $ 2^6 7^2 = 3136 $, $ 2^6 11^2 = 7744 $, and $ 2^6 13^2 = 10816 $. If $ p=3 $, then $ k = 729q^2 $. So $ 2006 \le 729q^2 \le 12004 \Rightarrow 2.75\ldots \le q^2 \le 16.46\ldots $. For $ q $ an integer, this holds when $ 2 \le q \le 4 $. Since $ q $ is a prime distinct from $ p=3 $, we have $ q=2 $. So if $ p=3 $, $ k = 3^6 2^2 = 2916 $. If $ p \ge 5 $, then $ k \ge 15625q^2 > 12004 $, a contradiction. So we have found all possible values of $ k $. The sum of the possible values of $ n = k - 2005 $ is thus \begin{align*} &(3136-2005) \\ + &(7744-2005)\\ + &(10816-2005)\\ + &(2916-2005)\\ = &\boxed{16592}.\end{align*}