Let $a_1,a_2,\dots$ be integers such that $a_1-a_2+a_3-a_4+\dots +{(-1)}^{n-1}{a}_n=n$, for all $n\ge 1$. Then $a_{51}+a_{52}+\dots +a_{1023}$ equals

Given:
$a_{n+1}-a_n={(-1)}^{n}n$

Let's write down the terms:
$a_1=1$
$a_2=a_1-1=1-1=0$
$a_3=a_2+2=0+2=2$
$a_4=a_3-3=2-3=-1$
$a_5=a_4+4=-1+4=3$
$a_6=a_5-5=3-5=-2$

We can notice a pattern for the terms:
For even index $2k$:
$a_{2k}=1-k$
For odd index $2k+1$:
$a_{2k+1}=k+1$

Let's check consecutive terms $a_{2k+1}+a_{2k+2}$:
$a_{2k+1}+a_{2k+2}=(k+1)+(1-(k+1\left)\right)=(k+1)+(-k)=1$

The sum to evaluate is:
$S=a_{51}+a_{52}+a_{53}+a_{54}+...+a_{1022}+a_{1023}$
We can pair the terms from the beginning:
$S=(a_{51}+a_{52})+(a_{53}+a_{54})+...+(a_{1021}+a_{1022})+a_{1023}$

Each pair of the form $a_{2k+1}+a_{2k+2}$ equals $1$.
The number of such pairs from $51$ to $1022$ is:
$\frac{1022-51+1}{2}=\frac{972}{2}=486$ pairs.

So, the sum of pairs is $486\cdot 1=486$.

Now we add $a_{1023}$:
Since index is odd, $1023=2k+1\Rightarrow 2k=1022\Rightarrow k=511$.
$a_{1023}=k+1=511+1=512$.

But we must check the alternating signs and the target sum options, as some conventions or calculations might yield 1 or 0 under different indexing or pair definitions.