No. Assume, for a contradiction, that such a permutation $a_1, \cdots , a_{2004}$ does exist. Then from $a_i + \cdots a_{i+9} = 0 \mod [10]$ and $a_{i+1} + \cdots + a_{i+10} = 0 \mod [10]$, we deduce that $a_i = a_{i+10} \mod [10]$ for all $i < 1995$. Since the $a_i$'s must give all the esidues modulo 10, it follows that $a_1, a_2, \cdots , a_{10}$ give all the residues modulo 10. Therefore, $a_1, \cdots ,a_{10}$ is a permutation of $1,2,3,4,5,6,7,8,9,10$ modulo 10. Thus $a_1 + \cdots + a_{10} = 1+2+ \cdots + 10 = 5 \mod [10]$, a contradiction. Pierre.

Note that if you replace "the sum of any 10 consecutive numbers is divisble by 10" by "the sum of any $n$ consecutive numbers is divisble by $n$" where $n$ is any odd number, then we can do this, and the indentity permutation works (as do many others). In particular, $a+(a+1)+\cdots+(a+n-1)=n\left(a+\frac{n-1}2\right).$ But Pierre's argument prohibiting such a permuation works just as well for any even number $n$ in place of 10, because if $n$ is even, $\frac{n(n+1)}2=\frac n2\cdot(n+1)$ cannot be divisible by $n$ because $n$ and $n+1$ are relatively prime.