Let arithmetic progression had integer means $a_i,a_j,i<j$ and $a_l$ is not integer when $l<j,l\not=i,0\le i,j,l \le 1998$. Then $j>2i$ and $a_l$ is integer if and only if $l=i+kd,d=j-i>i.$ Consider arithmetic progression $a_i=i/d, i=0,1,\dots 1998$ and $b_i=(i+1)/d,i=0,1,\dots 1998$. Exactly $1+[\frac{1998}{d}]$ terms $a_i$ are integer and exactly $[\frac{1999}{d}]$ terms of $b_i$ are integer. Therefore if d=29,30,... give arithmetic progression $a_i$ or $b_i$ with exactly n integer terms for n<70. If d=28 we have 71 or 72 integer terms. Therefore n=70 is solution.