Sum up

Answer: 223

Using nine consecutive numbers and estimation

Imagine nine consecutive numbers that add up to $2012.$

Their average is $2012\div9=223.\dot{5}$, so could be:

        ___  ___  ___  ___  223  ___  ___  ___  ___

or     ___  ___  ___  ___  224  ___  ___  ___  ___  

219 + 220 + 221 + 222 + 223 + 224 + 225 + 226 + 227 = 2007

Sum needs to increase by 5 to be 2012 so add 228 and remove 223

or

220 + 221 + 222 + 223 + 224 + 225 + 226 + 227 + 228= 2016

Sum needs to decrease by 4 to be 2012 so add 219 and remove 223

Using last digits

Any ten consecutive numbers will have last digits $0,1,2,3,4,5,6,7,8,9$ in some order.

$1+2+3+4+5+6+7+8+9=45$ so adding together all ten gives last digit of $5$

$2012$ last digit is $2$ so the number ending in $3$ is missing.

The numbers are all approximately $2012\div9=223.\dot{5}$ so $223$ must be the missing number.

Using algebra and multiples

Say the $10$ numbers are $n,$ $n+1,$ $n+2,$ $n+3,...,n+9$

Sum of all $10$ is $n+(n+1)+...+(n+9) = 10n+45$

If $n+r$ was removed, then $$\begin{align}10n+45 - (n+r)&=2012\\

\underbrace{9n+45}_{\text{multiple of 9}}-r&=2012\end{align}$$ The smallest multiple of $9$ after $2012$ is $2016$ so $9n+45=2016$ and $r=4$ (since $r\le9$). $$\begin{align}9n+45&=2016\\ \Rightarrow 9n &= 1971\\

\Rightarrow n & =219\end{align}$$ $n+r=219+4=223$ was removed.