[ Ordinary fractions ]  [ Inclusion-exclusion principle ]

How many irreducible fractions with numerator 2015 are there, which are less than $1 / 2015$ and greater than $1 / 2016$?
Let $a>0$ be the denominator of the desired fraction, then $\frac{1 / 2016 < 2015 / a < 1 / 2015}{2015^{2} < a < 2015 \cdot 2016} 2015^{2} < a$ $<2015^{2}+2015$. Therefore, the desired values of $a$ are numbers of the form $2015^{2}+n$, where $n$ is a natural number, $1 \leq n \leq$ 2014 and $\text{GCD}(2015; n)=1$.

Since $2015=5 \cdot 13 \cdot 31$, the values of $n$ that do not satisfy the condition are all natural numbers in the given interval that are divisible by at least one of the prime factors in this factorization. Let's find their quantity:

The multiples of 5 will be $13 \cdot 31-1=402$ (since the number 2015 itself is subtracted), the multiples of 13 will be 5$\cdot$31 - 1 = 154, and the multiples of 31 will be $5 \cdot 13-1=64$. At the same time, numbers that are multiples of two prime factors of the factorization will be counted twice (there are no numbers counted three times in the given interval). Thus, the numbers that are multiples of both 5 and 13 (i.e., multiples of $5 \cdot 13$) will be $31-1=30$, the numbers that are multiples of both 5 and 31 will be $13-1=12$, and the numbers that are multiples of both 13 and 31 will be $5-1=4$.

Therefore, the desired number of fractions is 2014 - $402-154-64+30+12+4=1440$.