XI APMO

We first note that the integer terms of any arithmetic progression are "equally spaced", because if the $i$th term $a_i$ and the $(i+j)$th term $a_{i+j}$ of an arithmetic progression are both integers, then so is the $(i+2j)$th term $a_{i+2j}=a_{i+j}+(a_{i+j}-a_i)$.

Thus, by scaling and translation, we can assume that the integer terms of the arithmetic progression are $1,2,\cdots ,n$ and we need only to consider arithmetic progression of the form $$1,1+\frac{1}{k},1+\frac{2}{k},\cdots ,1+\frac{k-1}{k},2,2+\frac{1}{k},\cdots ,n-1,\cdots ,n-1+\frac{k-1}{k},n$$

This has $kn-k+1$ terms of which exactly $n$ are integers. Moreover, we can add up to $k-1$ terms on either end and get another arithmetic progression without changing the number of integer terms.

Thus there are arithmetic progressions with $n$ integers whose length is any integer lying in the interval $[kn-k+1,kn+k-1]$, where $k$ is any positive integer. Thus we want to find the smallest $n>0$ so that, if $k$ is the largest integer satisfying $kn+k-1\le 1998$, then $(k+1)n-(k+1)+1\ge 2000$.

That is, putting $k=\lfloor 1999/(n+1)\rfloor$, we want the smallest integer $n$ so that $$\lfloor \frac{1999}{n+1}\rfloor (n-1)+n\ge 2000.$$

This inequality does not hold if $$\frac{1999}{n+1}\cdot (n-1)+n<2000$$

This simplifies to $n^2<3999$, that is, $n\le 63$. Now we check integers from $n=64$ on: $$\begin{array}{rl}& \text{for }n=64,\lfloor \frac{1999}{65}\rfloor \cdot 63+64=30\cdot 63+64=1954<2000;\\ & \text{for }n=65,\lfloor \frac{1999}{66}\rfloor \cdot 64+65=30\cdot 64+65=1985<2000;\\ & \text{for }n=66,\lfloor \frac{1999}{67}\rfloor \cdot 65+66=29\cdot 65+66=1951<2000;\\ & \text{for }n=67,\lfloor \frac{1999}{68}\rfloor \cdot 66+67=29\cdot 66+67=1981<2000;\\ & \text{for }n=68,\lfloor \frac{1999}{69}\rfloor \cdot 67+68=28\cdot 67+68=1944<2000;\\ & \text{for }n=69,\lfloor \frac{1999}{70}\rfloor \cdot 68+69=28\cdot 68+69=1973<2000;\\ & \text{for }n=70,\lfloor \frac{1999}{71}\rfloor \cdot 69+70=28\cdot 69+70=2002\ge 2000.\end{array}$$

Thus the answer is $n=70$.