Try $n=2$. You will easily see the pattern of the numbers given to each vertex. When $n=3$, you can divide the entire triangle into three triangles whose length of the side is 2. Apply the pattern we found into the case. Easy to see that every vertex should be given 0. $n\ge4$, divide the triangle into small triangles whose length of the side is 3.

Vexation wrote: Try $n=2$. You will easily see the pattern of the numbers given to each vertex. When $n=3$, you can divide the entire triangle into three triangles whose length of the side is 2. Apply the pattern we found into the case. Easy to see that every vertex should be given 0. $n\ge4$, divide the triangle into small triangles whose length of the side is 3. I think it works when $n=3$. $$0$$$$-1,+1$$$$+1,0,-1$$$$0,-1,+1,0$$

We firstly consider $n=2$. [asy][asy] size(2cm);defaultpen(fontsize(12pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; for(int a=0; a< 2+1; ++a) {for(int b=0; b< 2-a+1; ++b) {dot((a+b/2,b*0.86602540378));}} label("$a$",(0,0),N);label("$b$",(1,0),N); label("$c$",(2,0),N);label("$d$",(0.5,0.86602540378),N);label("$e$",(1.5,0.86602540378),N);label("$f$",(1,2*0.86602540378),N); [/asy][/asy] [asy][asy] size(2cm);defaultpen(fontsize(12pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; for(int a=0; a< 2+1; ++a) {for(int b=0; b< 2-a+1; ++b) {dot((a+b/2,b*0.86602540378));}} label("$a$",(0,0),N);label("$b$",(1,0),N); label("$c$",(2,0),N);label("$c$",(0.5,0.86602540378),N);label("$a$",(1.5,0.86602540378),N);label("$b$",(1,2*0.86602540378),N); [/asy][/asy] Note that we have $5$ different suitable triangles, hence we must have $$\begin{cases} a+b+c=0\\ d+e+f=0\\ a+b+d=0\\ b+c+e=0\\ d+e+f=0.\\ \end{cases}$$Solving this, we obtain that $a=d,b=e,c=f$ with condition $a+b+c=0$. Now consider $n=3$. Due to $n=2$ case, we have all three vertices of big triangle the same, hence they are zeros. Easy to examine that we get the following triangle, [asy][asy] size(4cm);defaultpen(fontsize(12pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; for(int a=0; a< 3+1; ++a) {for(int b=0; b< 3-a+1; ++b) {dot((a+b/2,b*0.86602540378));}} label("$0$",(0,0),N);label("$x$",(1,0),N); label("$-x$",(2,0),N);label("$0$",(3,0),N);label("$-x$",(0.5,0.86602540378),N);label("$0$",(1.5,0.86602540378),N);label("$x$",(2.5,0.86602540378),N); label("$x$",(1,2*0.86602540378),N);label("$-x$",(2,2*0.86602540378),N);label("$0$",(1.5,3*0.86602540378),N); [/asy][/asy] For $n=4$, note that we have three $n=3$ triangles in it, hence this overlapping essentially implies that all vertices must own $0$. For $n\geq 5$, we use one less overlapping triangles, hence here also all vertices own $0$.