We start playing with the condition, and sooner or later, we notice that the sum of vectors in the opposite nodes of the rhombuses are equal. This motivates us to generalize the condition: Lemma. Denote by $f(P)$ the number assigned to node $P$. Then for any subset of nodes $P_1,P_2,\dots,P_k$ with position vectors $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_k$ satisfying $\sum_{i=1}^k q_i\mathbf{v}_i=0$ for some real numbers $q_1,\dots,q_k$ with sum zero, we have $$\sum_{i=1}^k q_i f(P_i)=0.$$ (Note that the position vector condition is independent of the coordinate system because the sum of the $q_i$'s is zero.) Proof. We use the condition to start eliminating the points until this whole thing reduces to the $n=2$ case. (Essentially, this is induction on $k$.) Let's demonstrate how to eliminate the numbers assigned to the nodes on side $BC$ (this shows us how to eliminate the numbers row by row). Let the points on $BC$ be $B=X_0,X_1,\dots,X_n=C$ in this order, the next row be $Y_1,Y_2,\dots,Y_n$ and the row after that be $Z_1,Z_2,\dots,Z_{n-1}$. Then we can first eliminate $B$ like $f(X_0)=f(X_1)+f(Y_1)-f(Y_2)$ and $C$ in the same fashion, and after that eliminate all the $f(X_i)$ via $f(X_i)=f(Y_i)+f(Y_{i+1})-f(Z_i)$ for $1\le i\le n-1$. This clearly works if we have $\ge 3$ rows, so suppose we have reduced everything to a single equilateral triangle $XYZ$. But suppose that some reals $x+y+z=0$ satisfy $x\vec{X}+y\vec{Y}+z\vec{Z}=0$, then we have $z=-x-y$, hence this rearranges to $x\overrightarrow{ZX}+y\overrightarrow{ZY}=0$, which clearly can only hold for $x=y(=z)=0$ as $\overrightarrow{ZX}$ and $\overrightarrow{ZY}$ are linearly independent. So in this reduced case, $xf(X)+yf(Y)+zf(Z)=0$ is trivial. This proves our claim. $\blacksquare$ It is easy to finish from here. Indeed, if nodes $P,Q,R$ are such that $R$ is on segment $\overline{PQ}$, then $\vec{R}=\lambda \vec{P}+\mu \vec{Q}$ for $\lambda,\mu>0$ and $\lambda+\mu=1$, hence by the Lemma, $f(R)=\lambda f(P)+\mu f(Q)$, which is between $f(P)$ and $f(Q)$. Hence, every number assigned to all the nodes other than $A,B,C$ are strictly between the minimum and the maximum of $a,b,c$, so the answer in 1) is $1$, the diameter of $\triangle ABC$. (This assumes $a,b,c$ are pairwise distinct; if they need not be, then part 1) is trivially $\frac1n$.) For part 2), note that $\triangle ABC$ has rotational symmetry around its center $O$. If the equilateral $\triangle PQR$ also has center $O$, then $\vec{P}+\vec{Q}+\vec{R}=\vec{A}+\vec{B}+\vec{C}$ holds (it is enough to check this in the coordinate system around $O$), hence $f(P)+f(Q)+f(R)=a+b+c$ by the Lemma. Therefore if the total number of nodes is $N$, then the answer is $N\cdot \frac{a+b+c}3$. To compute $N$, just recall that by construction, $N=1+2+\dots+(n+1)=\frac{(n+1)(n+2)}2$.