Solve (1 + 3 + 5 + ... + (2n - 1))/(2 + 4 + 6 + ... + 2n) = 2017/2018.
(1 + 3 + 5 + ... + (2 n - 1)) = sum_(k=1)^n(-1 + 2 k) = n^2
(2 + 4 + 6 + ... + 2 n) = sum_(k=1)^n 2 k = n (n + 1)
n = 2017
