Solution:

Computing terms, $a_3 = 1000 - x,$ $a_4 = 2x - 1000,$ $a_5 = 2000 - 3x,$ $a_6 = 5x - 3000,$ and in general $$a_{2k+1} = 1000 F_{2k-1} - x F_{2k}, \qquad a_{2k+2} = x F_{2k+1} - 1000 F_{2k},$$ where $F_1 = F_2 = 1,$ $F_3 = 2, \ldots$ are the Fibonacci numbers. The sequence keeps going exactly as long as its terms stay nonnegative, so a long sequence requires $\frac{x}{1000}$ to be squeezed between the ratios $\frac{F_{2k}}{F_{2k+1}}$ and $\frac{F_{2k-1}}{F_{2k}}$ for larger and larger $k.$

For the first $13$ terms to be nonnegative we need $a_{12} = 89x - 55000 \ge 0$ and $a_{13} = 89000 - 144x \ge 0,$ i.e. $617.9\ldots \le x \le 618.05\ldots,$ so $x = 618.$ If $x \le 617$ the sequence turns negative by $a_{12},$ and if $x \ge 619$ it turns negative by $a_{13},$ so every other integer gives a shorter sequence.

Indeed $x = 618$ yields $1000, 618, 382, 236, 146, 90, 56, 34, 22, 12, 10, 2, 8, -6,$ a sequence of $14$ terms, the maximum possible. The answer is $618.$