SAUDI ARABIAN MATHEMATICAL COMPETITIONS

First, we notice that for all $2\le k\le 2017$, we always can find a partition satisfy i). For example, we can choose $a_i=2i-1$, $b_i=2i$ with $i=1,2,\dots ,1009$.

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 $k\equiv 1(\mathrm{mod}5)$ satisfy the given condition.

Indeed, we have some cases:

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

2. If $k\equiv 6(\mathrm{mod}10)$ 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 $k\equiv 1(\mathrm{mod}5)$ 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 $T=1008+k\not\equiv 9(\mathrm{mod}10)$.

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 $T=1009+2(k-1)\not\equiv 9(\mathrm{mod}10)$.

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