General Tso was leading a phalanx of chickens and knew there were between 1000 and 2000 in all. He desired a more accurate estimate, and so had a lieutenant line them up in columns of 11, 13, and 17 and use the Chinese Remainder Theorem to obtain a good tally for the general. The general was very happy with the resulting numbers until he found out that before the first two assemblies, Chicken Little was still sleeping. If Chicken Little made the third assembly (and was the only chicken still in the coop), how wrong were General Tso’s “more accurate” numbers?
The wording of the question is a little difficult, but the way I am interpreting this question is that the chickens are lined up in 3 columns/assemblies. The first column/assembly has 11 chickens, the second column/assembly has 13 chickens, and the third column/assembly has 17 chickens. One paricular chicken does not show up in the lineup for 11 or 13 but showed up for the assembly of 17.
So this is what I think the system of congruences will look like:
$x$ $\equiv$ 1 mod 11
$x$ $\equiv$ 1 mod 13
$x$ $\equiv$ 0 mod 17
Then, we solve for the system of congruences.
$x = 17a + 0$
$17a + 0 \equiv 1 \ (mod \ 13)$
$4a \equiv 1 \ (mod \ 13)$
$40 a \equiv 10 \ (mod \ 13) $
$a \equiv 13b + 10 $
$x = 17(13b + 10) = 221b + 10$
$x \equiv 1 \ (mod \ 11)$
$221b = 10 = 1 \ (mod \ 11)$
$b + 10 = 1 \ (mod \ 11)$
$b = -9 \ (mod \ 11)$
$b = 2 \ (mod \ 11)$
$b = 11c + 2$
$221(11c + 2) + 10 = 2431c + 452$.
Did I do something wrong in my calculations, because my remainder is not between 1000 and 2000. Or is my entire setup incorrect?
I think you're misunderstanding the problem.
Tso had his lieutenant line the chickens up in rows of 11, 13, and 17 chickens, and in each case note down how many chickens there were in the last, partial, row. This gave Tso three numbers $a$, $b$, $c$, and Tso then solved the congruence $$ x\equiv a \pmod{11} \qquad\qquad x\equiv b \pmod{13} \qquad\qquad x\equiv c \pmod{17} $$
Note that we're not told what $a$, $b$ and $c$ was, so we can't know what Tso's initial result was.
What we are told is the $a$ and $b$, but not $c$, were each $1$ too small. We're then asked to figure out how wrong that initial result was.
You're solving the right congruence to find that difference. As noted in comments you must have made an arithmetic error somewhere, because the solution is $x\equiv 1717\pmod{2431}$.
So Tso's answer was $1717$ chickens too low -- but that can't have been true, because he knew there were between 1000 and 2000 chickens in the troop and he was happy with his answer.
Instead the initial answer must have been $2431-1717=714$ chickens too high. That way it can have been possible both for the truth and for the false initial answer to be between $1000$ and $2000$.