Peculiar mod

$(2k-1) | (n-k) \Leftrightarrow (2k-1)(b-1) = n-k \Leftrightarrow (2k-1)(2b-1) = 2n-1$ for some $b \ge 1$ . So $f(n)$ equals the number of positive divisors of $2n-1$ . It's odd if and only if $2n-1$ is a square. For $n \le 2014$ it can be $1^2, 3^2, \ldots, 63^2$ . There are $\fbox{32}$ such $n$ .

n has to be of the form $2(x^{2}-x)+1$ for x= 1,2,3,....

for n ≤2014 x can be upto 32.