Keep calm and sum up!

We pair up every number $i$ with the $17-i$ numbers greater than it and add all such products.

Let $T(n) = \displaystyle \sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

$S = 1.2 + 1.3 + \ldots + 1.17$

$\qquad + 2.3 + 2.4 + \ldots + 2.17$

$\qquad + \ldots$

$\qquad + 16.17$

$\qquad + 17.0$

$S = 1(1+2+\ldots+17 - 1)$

$\qquad + 2(1+2+\ldots+17 - 1-2)$

$\qquad + \ldots$

$\qquad + 16( 1+2+\ldots+17 - 1-2-\ldots - 16)$

$\qquad + 17(1+2+\ldots+17 - 1 - 2-\ldots - 16 - 17)$

$S = \displaystyle \sum_{r=1}^{17} r( T(17) - T(r))$

$S = T(17)\times T(17) - \frac{1}{2} \bigg( \displaystyle \sum_{r=1}^{17} r^3 + r^2 \bigg)$

$S = (T(17))^2 - \frac{(T(17))^2}{2} - \frac{(17)(18)(35)}{6}$

$S= \boxed{10812}$

OR

We can team up every number with other 16 numbers. In this way every product is added twice. So we divide it by 2.

$S = \frac{1}{2} \times \bigg( \displaystyle \sum_{r=1}^{17} r(T(17)-r) \bigg)$

$S = \frac{1}{2} \times \bigg( (T(17))^2 - \frac{(17)(18)(35)}{6} \bigg)$

$S = \frac{21624}{2} = \boxed{10812}$