Romanian Mathematical Olympiad

From $a+b=c+d$ we get $(a+b{)}^2=(c+d{)}^2$, hence $ab=cd$ and furthermore $a^2-2ab+b^2=c^2-2cd+d^2$. As $(a-b{)}^2=(c-d{)}^2$, we have $|a-b|=|c-d|$, which implies $a-b=c-d$ or $a-b=d-c$. Recall that $a+b=c+d$, so either $a=c,b=d$ or $a=d,b=c$. The numbers must have one of the forms $\overline{aaaa}$, $\overline{abba}$ or $\overline{abab}$, with $a\ne 0$ and $a\ne b$. In all, there are $9+9\cdot 9+9\cdot 9=9+81+81=171$ numbers.