hmm, try induction

Note that a positive integer can be expressed as a sum of $2k$ distinct positive integers less than or equal to $2000$ if and only if it is between $1+2+\dots+2k=k(2k+1)$ and $2000+1999+\dots+(2001-2k)=k(4001-2k)$ inclusive, where $k$ is a positive integer not greater than $1000$. To show that all integers from $1(2(1)+1)=3$ to $999(4001-2(999))=200097$ are possible values, we need to show that $(k+1)(2(k+1)+1) \le k(4001-2k)$ for all integers from $1$ to $998$ inclusive. This is equivalent to $2k^2+5k+3 \le -2k^2+4001k \iff 4k(999-k) \ge 3$ which is true for all integers from $1$ to $998$ inclusive. Note that $2001000=1+2+\dots+2000$ is also a possible value of $n$. It is obvious that $1$ and $2$ cannot be values of $n$. Since $999(4001-2(999))<200098, 200099<1000(2(1000)+1)$, $200098$ and $200099$ cannot be values of $n$. Therefore, the only positive integers which cannot be values of $n$ are $\boxed{1, 2, 200098, 200099}$.

This is KJMO 2020 P1 as long as I remember

This is more of an AIME-style proof There must be at least 2 numbers added to 0, or 2 numbers subtracted from 1+2+…+2000 = 2001*1000= 2001000. At least this will be add or subtract 1+2 = 3. Therefore the numbers 1, 2, 2000009 and 2000008 are unattainable. We show that these are the only numbers unattainable. Say we include 2k numbers in the sum. Then the minimum possible value is 1+….+ 2k = k(2k+1) and the maximum value is 2000+….+2000-(2k-1) = (4001-2k)(k). Lemma 1: We show that for all possible values of k+1, (k+1)(2k+1)< (4001-2k)k. This is of course true for all k+1 (quadratics inequality). This proves that everything is within bound, since the previous argument shows that the bounds as k increases, it is continuous and skips no values. Therefore 1, 2, 2000008, 2000009 are the only unachievable values.