Problem 2016

First, we notice that for all $2 \leq k \leq 2017$, we always can find a partition satisfy i). For example, we can choose $a_{i}=2i-1$, $b_{i}=2i$ with $i=1,2, \ldots, 1009$.

Denote $m$ as the number of pairs $(a_{i}, b_{i})$ with $|a_{i}-b_{i}|=k$ then
$$T=(1009-m) \cdot 1 + m \cdot k = 1009 + (k-1)m.$$
Hence, the necessary condition of $k$ is $(k-1)m$ is divisible by $10$. First, we can see that all numbers $k \equiv 1 \pmod{5}$ satisfy the given condition.

Indeed, we have some cases:

1. If $k \equiv 1 \pmod{10}$ then $10 \mid k-1$ and ii) is satisfied.

2. If $k \equiv 6 \pmod{10}$ then $k$ is even. Thus, we need to prove $m$ is even. This is true because we can write $m = m_{1} + m_{2}$ with $m_{1}, m_{2}$ are the number of pairs with the difference is $k$ that have same parity and different parity. In $X$, the number of even and odd numbers are equal and all the pairs $(a_{i}, b_{i})$ with $|a_{i}-b_{i}|=1$ also contain exactly $1$ number in each type of even/odd. These imply that $m_{1}=m_{2}$ or $m$ is an even number.

Hence, all numbers $k \equiv 1 \pmod{5}$ satisfy the given condition.

Next, we will prove that all other numbers $k$ do not satisfy the given condition.

3. If $k$ is odd then $k-1$ is not divisible by $5$. We choose the partition with $a_{1}=1$, $b_{1}=k+1$ and the rest are divided into pairs with $a_{i}, b_{i}$ being two consecutive numbers. It is easy to check that this partition satisfies i) but not ii) because $T=1008+k \not\equiv 9 \pmod{10}$.

4. If $k$ is even, $k-1$ is not divisible by $5$ then $k-1$ is an odd number that is not divisible by $5$. We choose the partition with exactly two pairs $(1, k+1), (2, k+2)$ and the rest are divided into pairs with $a_{i}, b_{i}$ being two consecutive numbers. Similarly with the previous case, we have $T=1009+2(k-1) \not\equiv 9 \pmod{10}$.

Therefore, the necessary and sufficient condition of $k$ is $k \equiv 1 \pmod{5}$. From $2$ to $2017$, we have $403$ numbers in total satisfy it.