Consider the expansion of $f(x)=(1 + x)^{1000}$ using the binomial theorem:
$$f(x)=(1 + x)^{1000} = \binom{1000}{0} + \binom{1000}{1}x + \\
\binom{1000}{2}x^2 + \binom{1000}{3}x^3 + \binom{1000}{4}x^4 + \ldots + \binom{1000}{1000}x^{1000}$$
We evaluate $f(x)$ at $i, i^2=-1,i^3=-i,i^4=1$, respectively.
\begin{align}
f(i) &= 1\cdot\left[ \binom{1000}{0}+\binom{1000}{4}+\dots+\binom{1000}{1000}\right] \\
&+i\cdot\left[ \binom{1000}{1}+\binom{1000}{5}+\dots+\binom{1000}{997}\right]\\
&-1\cdot\left[ \binom{1000}{2}+\binom{1000}{6}+\dots+\binom{1000}{998}\right]\\
&-i\cdot\left[ \binom{1000}{3}+\binom{1000}{7}+\dots+\binom{1000}{999}\right]\\
f(-1) &= 1\cdot\left[ \binom{1000}{0}+\binom{1000}{4}+\dots+\binom{1000}{1000}\right] \\
&-1\cdot\left[ \binom{1000}{1}+\binom{1000}{5}+\dots+\binom{1000}{997}\right]\\
&+1\cdot\left[ \binom{1000}{2}+\binom{1000}{6}+\dots+\binom{1000}{998}\right]\\
&-1\cdot\left[ \binom{1000}{3}+\binom{1000}{7}+\dots+\binom{1000}{999}\right]\\
f(-i) &= 1\cdot\left[ \binom{1000}{0}+\binom{1000}{4}+\dots+\binom{1000}{1000}\right] \\
&-i\cdot\left[ \binom{1000}{1}+\binom{1000}{5}+\dots+\binom{1000}{997}\right]\\
&-1\cdot\left[ \binom{1000}{2}+\binom{1000}{6}+\dots+\binom{1000}{998}\right]\\
&+i\cdot\left[ \binom{1000}{3}+\binom{1000}{7}+\dots+\binom{1000}{999}\right]\\
f(1) &= 1\cdot\left[ \binom{1000}{0}+\binom{1000}{4}+\dots+\binom{1000}{1000}\right] \\
&+1\cdot\left[ \binom{1000}{1}+\binom{1000}{5}+\dots+\binom{1000}{997}\right]\\
&+1\cdot\left[ \binom{1000}{2}+\binom{1000}{6}+\dots+\binom{1000}{998}\right]\\
&+1\cdot\left[ \binom{1000}{3}+\binom{1000}{7}+\dots+\binom{1000}{999}\right]\\
\end{align}
Let
\begin{align}
a&=\left[ \binom{1000}{0}+\binom{1000}{4}+\dots+\binom{1000}{1000}\right]\\
b&=\left[ \binom{1000}{1}+\binom{1000}{5}+\dots+\binom{1000}{997}\right]\\
c&=\left[ \binom{1000}{2}+\binom{1000}{6}+\dots+\binom{1000}{998}\right]\\
d&=\left[ \binom{1000}{3}+\binom{1000}{7}+\dots+\binom{1000}{999}\right]\\
\end{align}
where $b$ is our desired result. We have
\begin{align}
a+ib-c-id&=f(i) \\
a-b+c-d&=f(-1)\\
a-ib-c+id&=f(-i)\\
a+b+c+d&=f(1)\\
\end{align}
Solving these linear equations, we get
\begin{align}
b &= \frac{1}{4}\left[-i\cdot f(i)-f(-1)+i\cdot f(-i)+f(1)\right]\\
&= \frac{1}{4}\left[-i\cdot (1+i)^{1000}+i\cdot (1-i)^{1000}+2^{1000}\right]\\
\end{align}
where we have $(1+i)^2=2i,(1-i)^2=-2i$.
So
$$(1+i)^{1000}=(1-i)^{1000}=2^{500}$$
Finally, $$b = \frac{1}{4}\cdot 2^{1000}=2^{998}$$