14. Given $a, b, c, d$ (where $a<b<c<d$ ) are positive integers not exceeding 2012, and they form an arithmetic sequence in this order. How many different sets of possible values of $(a, b, c, d)$ are there?
(2 marks)
已知 $a 、 b 、 c 、 d$ （其中 $a<b<c<d ）$ 均為不超過 2012 的正整數, 且它們按此順序組成一個等差數列。問 $(a, b, c, d)$ 有多少組不同的可能值？
(2 分)
14. 673685
14. Let $x$ be the common difference of the arithmetic sequence $a, b, c, d$. We have $d=a+3 x$. The question is thus equivalent to counting the number of positive integer solutions to the equation $(a+3 x)+t=2013$.
When $x=1$, the equation becomes $a+t=2010$ and there are 2009 solutions (corresponding to $a=1,2, \ldots, 2009$ ). When $x=2$, the equation becomes $a+t=2007$ and there are 2006 solutions. Likewise, when $x=3,4,5, \ldots, 669,670$, there are $2003,2000,1997, \ldots, 5,2$ solutions respectively. It follows that the answer is
$$
2009+2006+\cdots+2=\frac{(2009+2)(670)}{2}=673685 .
$$