imc

We take cases on the thousands digit, which must be either $1$ or $2$: If the number is of the form $\overline{1bcd},$ where $b,c,d$ are digits, then we must have $d=1+b+c.$ Since $d\le 9,$ we must have $b+c\le 9-1=8.$ By casework on the value of $b$, we find that there are $1+2+\cdots +9=45$ possible pairs $(b,c)$, and each pair uniquely determines the value of $d$, so we get $45$ numbers with the given property. If the number is of the form $\overline{2bcd},$ then it must be one of the numbers $2000,2001,\dots ,2012.$ Checking all these numbers, we find that only $2002$ has the given property. Therefore, the number of numbers with the property is $45+1=\boxed{46}$ .