positive integers on the circle we have $101$ positive integers with sum $300$ arranged on the circle.
prove that there exists an arc such that all numbers on it have sum $200$.
probably Dirichlet's Pigeonhole principle should be used
 A: We have $ \sum_{i=1}^{101} a_i =300 $ 
Consider these sums: $a_1 , a_1+a_2, \dots , a_1+a_2+ \dots a_{100} $ 
If any of those sums is equal to $100$ then we are done because if, for example, for some $j, 1 \leq j \leq 100 $ sum $ a_1+a_2+ \dots a_j = 100  $ then $a_{j+1}+a_{j+2}+ \dots a_{101} =200 $
If that is not the case then consider these sums $\pmod {100} $ .
Since we have 100 sums two of these must have same remainder modulo $100$ (that's because in this case none of the sums is $\equiv 0 \pmod {100}$ so we have 99 possible remainders ) 
Let those two sums be 
$a_1+a_2+ \dots + a_i \equiv k \pmod {100} $
$ a_1+a_2+ \dots a_j \equiv k \pmod {100} $ where $j>i$
If we subtract these two we get 
$a_{i+1}+a_{i+2}+ \dots a_j \equiv 0 \pmod {100} $ 
this sum can't be equal to zero because  $a_i > 0 , \forall i $ so it has to be either $100$ or $ 200$.
If it's  $200$, we are done. 
if it's $100$ then the arc we are looking for is the arc NOT containing consecutive points $a_{i+1}, a_{i+2}, \dots a_j$ .
