jmc

Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$, then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$. 6. Consider $c\in \{0,1,2,3,4\}$. $1abc+1ab(c+1)$ has no carry if $a,b\in \{0,1,2,3,4\}$. This gives $5^3=125$ possible solutions. With $c\in \{5,6,7,8\}$, there obviously must be a carry. Consider $c=9$. $a,b\in \{0,1,2,3,4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b=9$, $a\in \{0,1,2,3,4,9\}$ have no carry. Thus, the solution is $125+25+6=\boxed{156}$.