jmc

We can write 1234567890 as $$12\cdot 10^8+34\cdot 10^6+56\cdot 10^4+78\cdot 10^2+90.$$ Note that $$10^8-1=99999999=99\cdot 1010101,$$ is divisible by 99, so $12\cdot 10^8-12$ is divisible by 99.

Similarly, $$\begin{array}{rl}10^6-1& =999999=99\cdot 10101,\\ 10^4-1& =9999=99\cdot 101,\\ 10^2-1& =99=99\cdot 1\end{array}$$ are also divisible by 99, so $34\cdot 10^6-34$, $56\cdot 10^4-56$, and $78\cdot 10^2-78$ are all divisible by 99.

Therefore, $$12\cdot 10^8+34\cdot 10^6+56\cdot 10^4+78\cdot 10^2+90-(12+34+56+78+90)$$ is divisible by 99, which means that $1234567890$ and $12+34+56+78+90$ leave the same remainder when divided by 99.

Since $12+34+56+78+90=270=2\cdot 99+72$, the remainder is $\boxed{72}$.