1. The number of positive integers $n$ that make $1^{2 \times 15}+2^{2005}+\cdots+n^{205}$ divisible by $n+2$ is
$=1.0$.
Let $S_{n}=1^{2008}+2^{2000}+\cdots+n^{2005}$, then
$$
S_{n}=n^{2008}+(n-1)^{2005}+\cdots+1^{2008} \text {. }
$$

By misalignment addition, we get
$$
\begin{aligned}
2 S_{n}= & 2+\left(2^{2005}+n^{2005}\right)+\left[3^{2005}+(n-1)^{2000}\right]+ \\
& \cdots+\left(n^{2005}+2^{2005}\right) .
\end{aligned}
$$

From $k^{20 \infty}+(n+2-k)^{2005} \equiv 0(\bmod (n+2))(k \geqslant 2)$, we know $2 S_{n} \equiv 2(\bmod (n+2))$, which means $S_{n}(n \geqslant 2)$ cannot be divisible by $n+2$.

Also, $S_{1}$ cannot be divisible by $n+2$, so, $S_{n}$ cannot be divisible by $n+2$.