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This proves the entire statement of L219 at author-recorded tier 0. Say that distinct $u,w\in V$ meet at level $h$ when $p_uh-s_u=p_wh-s_w$.

If $p_u=p_w$, meeting at any level forces $s_u=s_w$, which the distinct pairs exclude; so vertices with the same prime never meet. If $p_u\ne p_w$, they meet at most at one level, since meeting at levels $h\ne h'$ gives $(p_u-p_w)(h-h')=0$.

Fix $v\in V$. For each $h\le K$ the hypothesis at $m=p_vh-s_v$ says that the weights of the vertices $w$ with $p_wh-s_w=m$ sum to zero. This set contains $v$, and $c_v\ne0$, so it contains some $u_h\ne v$, which meets $v$ at level $h$. By the previous paragraph, $p_{u_h}\ne p_v$, and the vertices $u_1,\dots,u_K$ are distinct.

Suppose $p_{u_h}=p_{u_{h'}}=p'$ for some $h\ne h'$. Subtracting $p_vh-s_v=p'h-s_{u_h}$ from $p_vh'-s_v=p'h'-s_{u_{h'}}$ gives

$$s_{u_h}-s_{u_{h'}}=(p_v-p')(h'-h).$$

The admissible $n$ satisfies $n\equiv s_{u_h}$ and $n\equiv s_{u_{h'}}$ modulo $p'$, so $p'$ divides $(p_v-p')(h'-h)$. Being prime, $p'$ divides $p_v-p'$, hence $p_v$, or it divides $h'-h$. The first is impossible because $p_v\ne p'$ are primes, and the second because $0<|h'-h|<K<p'$. Thus $p_v,p_{u_1},\dots,p_{u_K}$ are $K+1$ distinct primes, each greater than $K$, and the product of all distinct $p_w$ is at least their product, which exceeds $K^{K+1}$. $\square$