1. Given $n$ is a natural number, $n^{2}+4 n+2009$ can be expressed as the sum of the squares of four consecutive natural numbers. Then the sum of all $n$ that satisfy this condition is $\qquad$ .
2.1.712.
Let $n^{2}+4 n+2009$
$$
=(m-1)^{2}+m^{2}+(m+1)^{2}+(m+2)^{2} \text {, }
$$

where $m$ is a positive integer. Then
$$
(n+2)^{2}+2005=(2 m+1)^{2}+5 \text {, }
$$

which implies $(2 m+1)^{2}-(n+2)^{2}=2000$.
Factoring, we get
$$
(2 m+n+3)(2 m-n-1)=2000 \text {. }
$$

Since $2 m+n+3$ and $2 m-n-1$ have the same parity, and $2 m+n+3>2 m-n-1$, we have
$$
\begin{array}{l}
(2 m+n+3,2 m-n-1) \\
=(1000,2),(500,4),(250,8), \\
(200,10),(100,20),(50,40) . \\
\text { Hence }(m, n) \\
=(250,497),\left(\frac{251}{2}, 246\right),(64,119), \\
(52,93),\left(\frac{59}{2}, 38\right),(22,3) .
\end{array}
$$

Therefore, the sum of all $n$ that satisfy the condition is
$$
497+119+93+3=712 \text {. }
$$