Simple Adding

Note that $YYY=111Y$ . The sum of all the integers from $1$ to $X$ is $\frac{X(X+1)}{2}$ . Thus $\begin{aligned} \frac{X(X+1)}{2} &= 111Y \\ X^2 + X &= 222Y \\ X^2 + X - 222Y &= 0 \\ X&=\frac{-1\pm\sqrt{1+888Y}}{2} \end{aligned}$

This can only be an integer if $1+888Y$ is a perfect square. We know that $Y$ is an integer with $1\leq Y\leq 9$ . Thus $\begin{array}{c|c|c} Y & 1+888Y & \sqrt{1+888Y} \\ \hline 1 & 889 & \sqrt{889} \\ 2 & 1777 & \sqrt{1777} \\ 3 & 2665 & \sqrt{2665} \\ 4 & 3553 & \sqrt{3553} \\ 5 & 4441 & \sqrt{4441} \\ 6 & 5329 & 73 \\ 7 & 6217 & \sqrt{6217} \\ 8 & 7105 & 7\sqrt{145} \\ 9 & 7993 & \sqrt{7993} \end{array}$ Thus $Y=6$ and $X=\frac{-1\pm73}{2}$ . Taking the positive solution yields $X=\frac{72}{2} = \boxed{36}$ .