Geometriculus?

First, define $L_n$ as the $nth$ line drawn by the speaker, so that $L_1 = \overline{AB}, L_2 = \overline{BC}, L_3 = \overline{CD}, L_4 = \overline{DE},$ and so forth. Notice that $L_1 || L_5, \ L_2 || L_6, \ L_3 || L_7, \ and \ L_4 || L_8.$ More generally, $L_n || L_{n+4},$ so the set of lines parallel to $L_n$ will include all lines $L_m$ where $m \equiv n \pmod 4$ and $m \leq 59$ since the speaker drew only 59 lines. This now gives
$\gamma = \{L_1, L_5, L_9, ... , L_{57}\}, \ n(\gamma) = 15$
$\beta = \{L_2, L_6, L_{10}, ... , L_{58}\}, \ n(\beta) = 15$
$\Delta = \{L_3, L_7, L_{11}, ... , L_{59}\}, \ n(\Delta) = 15$
$\Theta = \{L_4, L_8, L_{12}, ... , L_{56}\}, \ n(\Theta) = 14$

so the quartic polynomial with the above numbers as its roots will be of the form $c(x-15)^3(x-14)$ where $c$ is a constant. For the purpose of this solution, let $c = 1$ so that the needed polynomial is $(x-15)^3(x-14) = x^4 - 59x^3 + 1305x^2 - 12825x + 47250.$ Differentiating this polynomial twice gives $\frac{d^2}{dx^2} (x-15)^3(x-14) = 12x^2 - 354x + 2610.$

Finally, the answer is obtained by adding $12 + 354 + 2610 + 24 = \boxed{3000}$