Could someone please post a solution?

I had all the ingredients for a solution as soon as I saw the problem, except the little observation at the end, that eluded me until now! For a polyhedron (or a planar map), Euler's formula $V + F = E + 2$ yields the known inequality $E \leq 3V - 6$. For conformity, here is a proof. Let $F_k$, $k\geq 3$ be the number of facets with $k$ sides. Then $F = F_3+F_4+\cdots + F_k + \cdots$, and by a double-counting argument $2E = 3F_3 + 4F_4 + \cdots + kF_k + \cdots$. It follows that $2E-3F = F_4 + 2F_5 + \cdots + (k-3)F_k + \cdots \geq 0$, hence $3(E+2) = 3V + 3F \leq 3V + 2E$, and so $E \leq 3V - 6$. The next step is to notice that if we can draw an extra edge (a diagonal of a facet), keeping the graph planar, then no matter how we colour this extra edge, the excentricities of the vertices do not decrease (either stay the same, or increase by $2$). Therefore it is enough to prove the problem when all facets have been triangulated; under that construction we will in fact now have $2E=3F$, $E = 3V-6$ and thus $F = 2V-4$. Thirdly, denote by $X_f$ the number of excentric angles of a facet $f$. Since each facet is a triangle, it follows $X_f \leq 2$, and then $\sum S_v = \sum X_f \leq 2F = 4V - 8$, meaning there exist (at least) two vertices $u$ and $w$ of excentricities $S_u, S_w < 4$. And now, the final step. In all configurations, for any vertex, its excentricity must be an even number (trivially proved, but somehow taking me a little long to realize); this means $S_u, S_w \leq 2$, and so $S_u + S_w \leq 4$. EDIT. In fact, we proved the stronger statement that there exist four vertices $a,b,c,d$ having $S_a + S_b + S_c + S_d \leq 8$; this is the maximum that can be said, since we can colour the edges of a tetrahedron $abcd$ such that $S_a = S_b = S_c = S_d = 2$.

nice problem! Define the excentricity of a face $A$, namely $T_A$, to be the number of excentric angles it has. Note that the excentricity of any vertex has to be even. Let the vertex set be $V$, the edge set be $E$, and the face set be $F$. It suffices to prove that $\sum_{v\in V}S_v\leq 4|V|-4$. Let $k=\frac{2|E|}{|F|}$. By Euler's formula, we have $|V|-|E|+|F|=2$ so $|F|=\frac{|V|-2}{\frac{k}{2}-1}$. Any triangle face must have excentricity at most $2$. Let $f_m$ be the number of faces with $m$ edges. We will prove that \[f_3\geq (4-k)|F|.\]For the sake of contradiction, assume the contrary. We have \begin{align*} k|F|&=2|E|\\ &=\sum_{m=3}^\infty mf_m\\ &\geq 3f_3+4(|F|-f_3)\\ &>4|F|-(4-k)|F|=k|F|, \end{align*}contradiction. Now \begin{align*} \sum_{v\in V}S_v&=\sum_{f\in F}T_f\\ &\leq 2f_3+\sum_{m=4}^\infty mf_m\\ &=\sum_{m=3}^\infty mf_m-f_3\\ &\leq (2k-4)|F|\\ &=4|V|-8 \end{align*}so we are done. $\square$