Not so tough

Let $n^2 - 19n + 99 = k^2$ , which by the Quadratic Formula yields $n = \frac{19 \pm \sqrt{19^2 - 4 (1)(99-k^2)}}{2} = \frac{19 \pm \sqrt{4k^2 - 35}}{2}.$ Now in order for n to be an integer value, we require the discriminant $4k^2 - 35$ to equal a perfect square. Hence,

$4k^2 - 35 = m^2 \Rightarrow 4k^2 - m^2 = 35 \Rightarrow (2k+m)(2k-m) = 5^{1}7^{1}.$

The integer divisor pairs of 35 include $(1,35); (5,7); (-1,-35); (-5,-7)$ , and we now set up the following systems of equations against these pairs:

$1) 2k - m = 5; 2k + m = 7 \Rightarrow \boxed{k = 3, m = 1}.$

$2) 2k - m = 1; 2k + m = 35 \Rightarrow \boxed{k = 9, m = 17}.$

$3) 2k - m = -7; 2k + m = -5 \Rightarrow \boxed{k = -3, m = 1}.$

$4) 2k - m = -35; 2k + m = -1 \Rightarrow \boxed{k = -9, m = 17}.$

Our required values for k are $\pm{3}$ and $\pm{9}$ , and substituting these values back into the original expression for n gives:

$k = \pm{3} \Rightarrow n = \frac{19 \pm 1}{2} \Rightarrow n = \boxed{9, 10}.$

$k = \pm{9} \Rightarrow n = \frac{19 \pm 17}{2} \Rightarrow n = \boxed{1, 18}.$

Finally, the required sum of such n is $1 + 9 + 10 + 18 = \boxed{38}.$