Rename $N$ to $n$. We show that the maximum possible weight of bamboo panda can get is $(n-1)^2$ kilograms. This is clear for $n = 1$, since $M = m$ for trees of size one. We assume $n \geq 2$ in what follows. First let us show that this is achievable. Panda builds the path graph on $n$ vertices, and labels the vertices $0$ to $n-1$ from left to right. Then for each $0 \leq i < n$ he writes the number $i^2$. Clearly $m = 0$ and $M = (n-1)^2$, whence $M-m = (n-1)^2$. Therefore it is sufficient to show that this tree satisfies the "average condition". For the terminal vertices, we have $0^2+1 \geq 1^2$ and $(n-1)^2+1 \geq (n-2)^2$. For a vertex $1 \leq i \leq n-2$ we have \[i^2+1 \geq \frac{2(i^2+1)}{2} = \frac{(i-1)^2+(i+1)^2}{2}\](this is infact an equality) which implies that the construction is valid. It now suffices to show that $M-m \leq (n-1)^2$. Root the tree at a vertex labelled $M$. If there are multiple such vertices, we select one arbitrarily. Let $n_u$ denote the size of the subtree of the $u$th vertex. Moreover, let $p_u$ denote the parent of vertex $u$. Note that $p_u$ is defined for all vertices $u$ except for the root. Finally, define $x_u$ as the real number written on vertex $u$. We prove the following claim: Claim: Let $u$ be any vertex other than the root. Then $x_{p_u}-x_u \leq 2n_u-1$. Proof: We prove this by backward inducting on the distance of the vertex from the root. Suppose we've to prove the result for vertex $v$ (which is not the root). If $v$ is a leaf, from the average condition we have \[x_v+1 \geq x_{p_v}\]which implies that $x_{p_v}-x_{v} \leq 1$. Since $n_v = 1$, this implies the bound for all leaves. Otherwise, suppose $v$ has $k$ children, labelled $v_1, v_2, \dots, v_k$, where $k \geq 1$. From the average condition we have \[x_v+1 \geq \frac{x_{p_v}+\sum \limits_{i} x_{v_i}}{k+1}\]From the inductive hypotheses on a vertex $v_i$ we know that $x_{v_i} \geq x_{v}-2n_{v_i}+1$. Thus \[x_v+1 \geq \frac{x_{p_v}+\sum \limits_{i} x_{v_i}}{k+1} \geq \frac{x_{p_v}+\sum \limits_{i} (x_{v}-2n_{v_i}+1)}{k+1} = \frac{x_{p_v}+kx_v - \sum \limits_{i} (2n_{v_i}-1)}{k+1}\]Now note that \[n_{v_1}+\dots +n_{v_k} = n_v-1\]by definition. Thus the above simplifies to \[x_v+1 \geq \frac{x_{p_v}+kx_v+k-2n_v+2}{k+1}.\]Subtracting one from both sides and multiplying with $k+1$ yields \[(k+1)x_v \geq x_{p_v}+kx_v-2n_v+1\]which finally simplifies to \[x_v \geq x_{p_v}-2n_v+1\]or that $x_{p_v}-x_v \leq 2n_v-1$, finishing the inductive step. Now suppose the vertices on the path from $M$ to $m$ are $M_0 = M, M_1, M_2, \dots, M_k = m$ for some $k \geq 1$, so that the parent of $M_i$ for $1 \leq i \leq k$ is $M_{i-1}$. From the claim, we know that for all $1 \leq i \leq k$ we have \[x_{M_{i-1}}-x_{M_i} \leq 2n_{M_i}-1\]Adding, we obtain that \[M-m = M_0-M_k \leq \sum \limits_{i = 1}^k (2n_{M_i}-1)\]Now note that $n_{M_i} \leq n-i$, since the subtree of $M_i$ does not contain $M_0, M_1, \dots, M_{i-1}$. As a result \[M-m \leq \sum \limits_{i = 1}^k (2(n-i)+1) \leq \sum \limits_{i = 1}^n (2(n-i)+1) = (n-1)^2\]as desired. To conclude, the maximum weight of bamboo Panda can get is $(n-1)^2$ kilograms, or in the original nomenclature: $(N-1)^2$. $\square$

Answer: $(N-1)^2$ Achieved by a path with vertices from 1 to $N$ where $i$th vertex contains number $(i-1)^2$ and vertices $i$ and $i+1$ are connected. Easy to check that all inequalities are, in fact, equalities, except $(N-1)^2 + 1 \geq (N-2)^2$, which is obvious. Now we will prove that it's maximum. Induct on $N$. Base cases are obvious. First, let's label the "levels" of the tree, where level 1 is the root, level 2 are all the vertices connected to root and so on. Let the vertex with value $M$ be the root - $v_r$ (if there are multiple, take any). Let $T$ be the subtree of some level 2 vertex - $v_{r_T}$ - that contains the vertex with $m$ value (if there are multiple such level 2 vertices, choose any). Notice that the numbers on vertices of $T$ also satisfy the condition, because, if $x$ is the number written on its root, $S_x$ the sum of numbers connected to it, and $d_x$ is the degree of it, then $x+1 \geq \frac{S_x}{d_x} \geq \frac{S_x - M}{d_x-1}$ , where the last inequality holds because $M$ is the largest number written. And other inequalities don't change if we delete $v_r$. So $x - m \leq (|T| - 1)^2 \leq (N-2)^2$ by induction hypothesis. Now we are left to show that $M-x \leq 2N-3$. For this, we prove a general claim. Claim: Suppose that $v$ and $u$ are connected with $u$ located 1 level lower than $v$, and $x_u$ and $x_v$ - numbers written on it, then $x_u \geq x_v - 2|T_u| + 1$ ($T_u$ is the subtree of $u$). We will prove this result inducting downwards on the "level" of vertex $u$. For vertices of the last level, $x_u+1 \geq x_v$, because they are only connected to $v$. Suppose now that a vertex $u$ is connected to vertices $u_1, u_2, \ldots, u_m$ on a lower level and vertex $v$ on a higher level. We know that $x_{u_i} \geq x_{u} - 2|T_{u_i}|+1$ by induction hypothesis. If $x_v > x_u + 2|T_u| - 1$ then, using $|T_u| = \sum |T_{u_i}| + 1$, $x_u + 1 \geq \frac{x_{u_1} + x_{u_2} + \ldots + x_{u_m} + x_v}{m+1} > \frac{(m+1)x_u + m + 2 - 1}{m+1} = x_u + 1$, which is a contradiction. Now use the claim on $v_r$ and $v_{r_T}$. We get $x \geq M - 2|T_{v_{r_T}}|+1 \geq M - 2N +3$, which is what we wanted.