IMO 1989 Problem 1

I got a weird construction.

Prove that the set $ \{1,2, \ldots, 1989\}$ can be partitioned into disjoint subsets $A_1,A_2,\dots,A_{117}$ such that

i.) each $ A_i$ contains $17$ elements

ii.) the sum of all the elements in each $ A_i$ is the same.

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Proof

To simplify notations, we place numbers in a matrix $A$ whose columns are subsets.

Step 1: reduce $[1989]$ to $[117\cdot3]=[351]$, because we could then extend $A$ vertically to include $$\begin{pmatrix}3\cdot117+1 & 3\cdot117+2 & \dots & 4\cdot117\\ 5\cdot117 & 5\cdot 117-1&\dots&4\cdot117+1\\5\cdot117+1 & 5\cdot117+2 & \dots & 6\cdot117\\ 7\cdot117 & 7\cdot 117-1&\dots&6\cdot117+1\\ \vdots & \vdots & \ddots & \vdots \\ 15\cdot117+1 & 15\cdot117+2 & \dots & 16\cdot117\\ 17\cdot117 & 17\cdot 117-1&\dots&16\cdot117+1\end{pmatrix}.$$

Step 2: shift $[351]$ to $\{0,\pm1,\dots,\pm175\}$.

Step 3: make the first $88$ columns $$\begin{pmatrix}-175 & -174 & \dots & -88 \\ 87 & 85 & \dots & -87 \\ 88 & 89 & \dots & 175\\ \end{pmatrix}$$ and leaves $\{0,\pm2,\dots,\pm86\}$, or equivalently $\{0,\pm1,\dots,\pm43\}$.

Step 4: make the next $22$ columns $$2\begin{pmatrix}-43 & -42 & \dots & -22 \\ 21 & 19 & \dots & -21 \\ 22 & 23 & \dots & 43\\ \end{pmatrix}$$ and leaves $\{0,\pm2,\dots,\pm20\}$, or equivalently $\{0,\pm1,\dots,\pm10\}$.

Step 5: make the next $4$ columns $$4\begin{pmatrix}-10 & -9 & -8 & -7 \\ 3 & 1 & -1 & -3 \\ 7 & 8 & 9 & 10\\ \end{pmatrix}$$ and leaves $\{0,\pm2,\pm4,\pm5,\pm6\}$.

Step 6: make the lat $3$ columns $$4\begin{pmatrix}-2 & 2 & 5 \\ -4 & 4 & 0 \\ 6 & -6 & -5\\ \end{pmatrix}$$ and we are done.