Solution:

For $x \gt 0,$ $f(x) \to +\infty$ both as $x \to 0^+$ (the numerator tends to $18 \cdot 72 \cdot 98 \cdot k \gt 0$) and as $x \to \infty,$ so $f$ attains a global minimum value $c$ on $(0, \infty).$ It is attained at exactly two points precisely when $f(x) - c \ge 0$ with two distinct positive double roots, i.e. $$(x - 18)(x - 72)(x - 98)(x - k) - cx = (x^2 - Sx + P)^2$$ where the roots of $x^2 - Sx + P$ are positive and distinct (so $S, P \gt 0$).

Matching coefficients of $x^3,$ $x^2,$ and the constant (the $x$-coefficient just determines $c$): $$2S = 188 + k, \qquad S^2 + 2P = 10116 + 188k, \qquad P^2 = 18 \cdot 72 \cdot 98 \cdot k = 127008k.$$ Substitute $k = 2t^2$ with $t \gt 0:$ then $S = 94 + t^2$ and $P = 504t.$ The middle equation becomes $(94 + t^2)^2 + 1008t = 10116 + 376t^2,$ i.e. $$t^4 - 188t^2 + 1008t - 1280 = 0,$$ which factors as $(t - 2)(t - 4)(t + 16)(t - 10) = 0.$

The positive roots $t = 2, 4, 10$ give $k = 2t^2 = 8, 32, 200$ (each indeed yields $S^2 \gt 4P,$ matching the problem's promise of exactly three values). The sum is $8 + 32 + 200 = 240.$