jmc

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 $p^{20}\ge 2^{20}>12004$ 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 $2006\le 729q^2\le 12004\Rightarrow 2.75\dots \le q^2\le 16.46\dots$. 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 $$\begin{array}{rl}& (3136-2005)\\ +& (7744-2005)\\ +& (10816-2005)\\ +& (2916-2005)\\ =& \boxed{16592}.\end{array}$$