Let S be the common value for the twelve sums; we will confine ourselves to the search for the minimum of S: the maximum will then be obtained by changing any value x with 20-x (thus S is replaced by 60-S)
Let C be the central value, and let V, I, E be the 6-components arrays giving respectively the values on Vertices, External and Internal nodes; with SumV, SumI, SumE we will denote the sums of the elements of V, I, E respectively.
The most obvious relations are:

• V[j]+I[j]+C=S, thus SumV+SumI=6(S-C);
• 2SumV+SumE=6S;
• SumV+SumI+SumE+C = 1+...+19 = 190.

Remark that the second relation implies that SumE must be even; then the first relation implies that also SumI must be even; then the third relation implies that also C must be even.
From V[j]+I[j]=S-C (j=1,2,...6) we see that the value S-C must be obtained 6 times; the minimum for S, because of C even, corresponds to C=2, V[j]+I[j]=17 (say: 3+14, 4+13, ..., 8+9); thus we need S > 18.

No other relations seeming to be significant, we will search through a small javascript program, starting with S = 19 ...