We prove that $n=4998$ works. Note that $k^2 - (k+1)^2 - (k+2)^2 + (k+3)^2 = 4$, so $k^2 - (k+1)^2 - (k+2)^2 + (k+3)^2 - (k+4)^2 + (k+5)^2 + (k+6)^2 - (k+7)^2 = 4-4 = 0$. So we can let $k = 2013,2021,2029,\ldots,4985$, leaving us with $4993^2 \pm 4994^2 \pm 4995^2 \pm 4996^2 \pm 4997^2 \pm 4998^2$. Note that $493^2 - 4994^2 - 4995^2 + 4996^2 - 4997^2 + 4998^2$ $= 4 + (4998^2-4997^2)$ $= 4 + (4998+4997)(4998-4997)$ $= 4+9995 = 9999$

Slightly easier, but based on the same thing, just use $x^2 - (x + 1)^2 - (x + 2)^2 + (x + 3)^2 = 4$, and now take $2013^2 + 2014^2 + 2015^2 + 2016^2 + 2017^2$ and then minus a bunch of those fours, $5072784$ of them to be precise. We can make infinitely many of these by alternating the plus fours and the minus fours.

Got a easier solution to this one.. Simply observe - $2014^2-2013^3=(2014+2013)(2014-2013)=4027$. Now, $9999-4027=5972$.. We see, $4\mid{5972}$.. (specifically $4.(1493)=5972$) $\implies$ we can travel from $4027$ to $5972$ in exactly $1493$ steps each of length $4$. We also observe that, $(x+3)^2-(x+2)^2-(x+1)^2+x^2=4$.. Therefore, we can arrange all the numbers in the series, after $2014$, in groups of $4$ such that each group results in a $+4$ donation to the sum.. (As proved above, there will be exactly $1493$ such 'groups').. Therefore, we conclude that $\exists n\in\Bbb{N}$ for which the above sum (with the $+$ and $-$ signs as shown in each group) is equal to $9999$. Q.E.D. $\square$