3. For a sequence of real numbers $x_{1}, x_{2}, \cdots, x_{n}$, define its "value" as $\max _{1 \leqslant i \leqslant n}\left\{\left|x_{1}+x_{2}+\cdots+x_{i}\right|\right\}$. Given $n$ real numbers, David and George want to arrange these $n$ numbers into a sequence with low value. On one hand, diligent David examines all possible ways to find the minimum possible value $D$. On the other hand, greedy George chooses $x_{1}$ such that $\left|x_{1}\right|$ is as small as possible. From the remaining numbers, he chooses $x_{2}$ such that $\left|x_{1}+x_{2}\right|$ is as small as possible, $\cdots \cdots$ At the $i$-th step, he chooses $x_{i}$ from the remaining numbers such that $\left|x_{1}+x_{2}+\cdots+x_{i}\right|$ is as small as possible. At each step, if there are multiple numbers that give the same minimum absolute sum, George arbitrarily chooses one. Finally, the value of the sequence he gets is $G$. Find the smallest constant $c$, such that for every positive integer $n$, every array of $n$ real numbers, and every sequence George can obtain, we have
$$
G \leqslant c D \text {. }
$$
3. $c=2$.

If the initial numbers given are $1, -1, 2, -2$, then David arranges these four numbers as $1, -2, 2, -1$, and George can get the sequence $1, -1, 2, -2$.
Thus, $D=1, G=2$.
Therefore, $c \geqslant 2$.
We now prove: $G \leqslant 2 D$.
Let the $n$ real numbers be $x_{1}, x_{2}, \cdots, x_{n}$, and assume David and George have already obtained their arrangements. Suppose David and George get the sequences $d_{1}, d_{2}, \cdots, d_{n}$ and $g_{1}, g_{2}, \cdots, g_{n}$, respectively.
$$
\begin{array}{c}
\text { Let } M=\max _{1<i \leqslant n}\left\{\left|x_{i}\right|\right\}, \\
S=\left|x_{1}+x_{2}+\cdots+x_{n}\right|, \\
N=\max \{M, S\} .
\end{array}
$$

Then $D \geqslant S$,
$D \geqslant \frac{M}{2}$,
$G \leqslant N=\max \{M, S\}$.
From equations (1), (2), and (3), we get
$$
G \leqslant \max \{M, S\} \leqslant \max \{M, 2 S\} \leqslant 2 D \text {. }
$$

In fact, by the definition of value, equation (1) holds.
For equation (2), consider an index $i$ such that $\left|d_{i}\right|=M$.
Thus, $M=\left|d_{i}\right|$
$=\left|\left(d_{1}+d_{2}+\cdots+d_{i}\right)-\left(d_{1}+d_{2}+\cdots+d_{i-1}\right)\right|$
$\leqslant\left|d_{1}+d_{2}+\cdots+d_{i}\right|+\left|d_{1}+d_{2}+\cdots+d_{i-1}\right|$
$\leqslant 2$.
Therefore, equation (2) holds.
Finally, we prove that equation (3) holds.
Let $h_{i}=g_{1}+g_{2}+\cdots+g_{i}$.
We use mathematical induction on $i$ to prove $\left|h_{i}\right| \leqslant N$.
When $i=1$, $\left|h_{1}\right|=\left|g_{1}\right| \leqslant M \leqslant N$, so the conclusion holds.
Notice that, $\left|h_{n}\right|=S \leqslant N$, so the conclusion also holds.
Assume $\left|h_{i-1}\right| \leqslant N$, and consider two cases.
(1) Assume that $g_{i}, g_{i+1}, \cdots, g_{n}$ do not have two terms with opposite signs. Without loss of generality, assume they are all non-negative.
Then $h_{i-1} \leqslant h_{i} \leqslant \cdots \leqslant h_{n}$.
Thus, $\left|h_{i}\right| \leqslant \max \left\{\left|h_{i-1}\right|,\left|h_{n}\right|\right\} \leqslant N$.
(2) In $g_{i}, g_{i+1}, \cdots, g_{n}$, there are positive and negative numbers, so there exists an index $j \geqslant i$ such that $h_{i-1} g_{j} \leqslant 0$.
By the definition of the sequence George gets,
$$
\begin{array}{l}
\left|h_{i}\right|=\left|h_{i-1}+g_{i}\right| \leqslant\left|h_{i-1}+g_{j}\right| \\
\leqslant \max \left\{\left|h_{i-1}\right|,\left|g_{j}\right|\right\} \leqslant N .
\end{array}
$$

Thus, the conclusion holds.
This shows that equation (3) also holds.