Find the smallest positive integer $n$ with the following property: There does not exist an arithmetic progression of 1993 terms of real numbers containing exactly $n$ integers.
and Marking Scheme:
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 $(i+2 j)$ th term $a_{i+2 j}=a_{i+j}+\left(a_{i+j}-a_{i}\right)$.

1 POINT for realizing that the integers must be "equally spaced".
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 $k n-k+1$ 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.

2 POINTS for noticing that the maximal sequence has an equal number of terms on either side of the integers appearing in the sequence (this includes the 1 POINT above). In other words, 2 POINTS for the scaled and translated form of the progression including the $k-1$ terms on either side.

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

4 POINTS for clarifying the nature of the number $n$ in this way, which includes counting the terms of the maximal and minimal sequences containing $n$ integers and bounding them accordingly (this includes the 2 POINTS above).

That is, putting $k=\lfloor 1999 /(n+1)\rfloor$, we want the smallest integer $n$ so that

$$
\left\lfloor\frac{1999}{n+1}\right\rfloor(n-1)+n \geq 2000
$$

This inequality does not hold if

$$
\frac{1999}{n+1} \cdot(n-1)+n<2000
$$

2 POINTS for setting up an inequality for $n$.
This simplifies to $n^{2}<3999$, that is, $n \leq 63$. Now we check integers from $n=64$ on:

$$
\begin{aligned}
& \text { for } n=64,\left\lfloor\frac{1999}{65}\right\rfloor \cdot 63+64=30 \cdot 63+64=1954<2000 ; \\
& \text { for } n=65,\left\lfloor\frac{1999}{66}\right\rfloor \cdot 64+65=30 \cdot 64+65=1985<2000 ; \\
& \text { for } n=66,\left\lfloor\frac{1999}{67}\right\rfloor \cdot 65+66=29 \cdot 65+66=1951<2000 ; \\
& \text { for } n=67,\left\lfloor\left\lfloor\frac{1999}{68}\right\rfloor \cdot 66+67=29 \cdot 66+67=1981<2000 ;\right. \\
& \text { for } n=68,\left\lfloor\frac{1999}{69}\right\rfloor \cdot 67+68=28 \cdot 67+68=1944<2000 ; \\
& \text { for } n=69,\left\lfloor\frac{1999}{70}\right\rfloor \cdot 68+69=28 \cdot 68+69=1973<2000 ; \\
& \text { for } n=70,\left\lfloor\frac{1999}{71}\right\rfloor \cdot 69+70=28 \cdot 69+70=2002 \geq 2000 .
\end{aligned}
$$

Thus the answer is $n=70$.
1 POINT for checking these numbers and finding that $n=70$.