Copyright © University of Cambridge. All rights reserved.
Answer: $227$
Algebra from the first number
Let the first number be $n$, so the nine numbers are:
$n$, $n+1$, $n+2$, $n+3$, $n+4$, $n+5$, $n+6$, $n+7$, $n+8$.
We are told that they all add up to $2007$, so
$n+n$$+$$1+n$$+$$2+n$$+$$3+n$$+$$4+n$$+$$5+n$$+$$6+n$$+$$7+n$$+$$8=2007$,
$9n+36=2007$,
$9n=1971$,
$n=219$.
If the smallest number is $219$, the largest will be $219+8=227$.
Algebra from the middle number
Let the middle number be $n$, so the nine numbers are:
$n-4$, $n-3$, $n-2$, $n-1$, $n$, $n+1$, $n+2$, $n+3$, $n+4$.
Summing these gives $9n=2007$ so $n=223$.
Therefore the largest number, $n+4$, is $227$.
Averages
The average of the nine numbers is $2007\div9=223$, so they are $219$, $220$,$...$, $226$, $227$.
So the last number is $227$.
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.