Quixotic quadruples

If the terms $a,b,c,d$ are $b-u,b,b+u,b+2u$ , then $\begin{aligned} (b-u)^3 + b^3 + (b+u)^3 & = \; (b + 2u)^3 \\ 3b^3 + 6bu^2 & = \; b^3 + 6b^2u + 12bu^2 + 8u^3 \\ b^3 - 3b^2u - 3bu^2 - 4u^3 & = \; 0 \\ (b - 4u)(b^2 + bu + u^2) & = \; 0 \end{aligned}$ so that $b =4u$ and hence we must have $(a,b,c,d) = (3u,4u,5u,6u)$ for some positive integer $u$ .

• We have $1867 \le a=3u \le 2008$ provided that $623 \le u \le 669$ .
• We have $1867 \le b=4u \le 2008$ provided that $467 \le u \le 502$ .
• We have $1867 \le c=5u \le 2008$ provided that $374 \le u \le 401$ .
• We have $1867 \le d=6u \le 2008$ provided that $312 \le u \le 334$ .

and so there are $(669 - 623 + 1) + (502 - 467 + 1) + (401 - 374 + 1) + (334 - 312 + 1) \; = \; 47 + 36 + 28 + 23 \; = \; 134$ choices for $u$ , so there are $\boxed{134}$ quixotic quadruples.