BUMP.////

As idea. We will prove for $n$ If $a_1=1$ then $-a_2^{-3}+...-a_n^{-3}=\frac{1}{2}$ - problem for $n-1$ So let there are not $a_i=1$ Let there are no more than $2$ same elements. Then $\frac{1}{2}=a_1^{-3}+...+a_n^{-3} \leq 2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{(n+1)^3})$ But $2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{(n+1)^3})<2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{(n+1)^3}+...)<2(\frac{1}{2^3}+\frac{1}{3^3}+...\frac{1}{15^3}+ 16*\frac{1}{16^3}+32*\frac{1}{32^3}+....)<2(\frac{1}{2^3}+\frac{1}{3^3}+...+\frac{1}{15^3}+\frac{3}{64})<\frac{1}{2}$ - contradiction

From RagvaloD, we may assume that all integers $a_i$ are greater than $1$. We now show that the maximum possible sum $2\sum_{n=2}^{1001}\frac{1}{n^3}$ must be less than $\frac{1}{2}$. Notice that $\sum_{n=a}^b \frac{1}{n^3} < \int_{a-1}^b \frac{1}{x^3} dx$. Then $$2\sum_{n=2}^{1001}\frac{1}{n^3} < \frac{2}{2^3} + 2\int_2^{1001} \frac{1}{x^3} dx=\frac{1}{4}-\frac{1}{x^2}\biggr|_2^{1001}=\frac{1}{4}+\frac{1}{4}-\frac{1}{1001^2}<\frac{1}{2}.$$ One possible set of integers is $a_k=2$ for $1\leq k \leq 4$ and $a_k=0$ for $5 \leq k \leq 2002$, and we are done. It is interesting to note that there must be three copies of the same number even when $i > 2002$. Also, the approximate sum in question rounds to about $0.4041$.