One hundred friends, including Alice and Bob, live in several cities. Alice has determined the distance from her city to the city of each of the other 99 friends and totaled these 99 numbers. Alice’s total is 1000 km. Bob similarly totaled his distances to everyone else. What is the largest total that Bob could have obtained? (Consider the cities as points on the plane; if two people live in the same city, the distance between their cities is considered zero).
The answer is $\boxed{99000}$. Call the other $98$ friends $C_1 ,C_2 , \ldots, C_{98}$ and let $a_i$ be the distance from Alice to $C_i$ and $b_i$ the distance between Bob and $C_i.$ Finally, denote the distance between Alice and Bob $r.$ Then the desired sum is just
$$r+b_1+b_2 +\cdots+ b_{98} \leq r+ (r+a_1)+(r+a_2)+\cdots+(r+a_{98})= 99 r+ a_{1}+\cdots+a_{98}=98r+1000 \leq 99000$$where the first inequality is by the triangle inequality and the last equality follows from $r\leq 1000.$ Equality can be reached if Alice and Bob live $1000$ kilometers apart and all other friends live in Alice's city.