Let $m = 1^2 + 2^2 + ... + 2001^2 + 2002^2 + 2003^2 + 2005^2$ $m$ is odd, so $m = 2n+1$ for some integer $n$. $m + n^2 = (n+1)^2$, and noting that $n > 2005$, the left side is the sum of $2005$ different positive square numbers.

my proof is exactly the same.

I though about this about a half of hour and found that the following is also true: There exist $0<a_1<a_2<\cdots<a_{2005}$ so that $\sum_{j=1}^ka_j^2$ are all squares for all $k=1,2,\ldots,2005$.

it suffices to notice that: $3^2+4^2=5^2$; $5^2+12^2=13^2$; $13^2+84^2=85^2$; ...... then it's trivial!