New Year's Countdown Day 28: Percentage Puzzle

Let $N$ be the integer we are looking for. Let $x$ and $y$ be positive integers such that $0.72x = 1.28y = N$ i.e. $N$ is 28% smaller than $x$ and 28% larger than $y.$ Normally, we would reduce the equality into simplest terms, but because we want $N$ to be an integer, we must take into consideration the decimal part:

$\begin{aligned} 0.72x &= 1.28y \\ \dfrac{72}{100}x &= \dfrac{128}{100}y \\ \dfrac{9}{100}x &= \dfrac{16}{100}y \\ \dfrac{9}{100}x &= \dfrac{4}{25}y. \end{aligned}$

We must have $100|x$ and $25|y$ in order for $N$ to be an integer. Let $x = 100x'$ and $y = 25y'.$ We have

$\begin{aligned} \dfrac{9}{100}(100x') &= \dfrac{4}{25}(25y') \\ 9x' &= 4y'. \end{aligned}$

We see that $4|x'$ and $9|y'.$ The smallest possible value of $N$ occurs when we minimize $x$ and $y.$ The minimum possible values of $x', y'$ are 4 and 9, respectively, so the minimum possible values of $x, y$ are

$\begin{aligned} x &= 100x' = 100(4) = 400 \\ y &= 25y' = 25(9) = 225. \end{aligned}$

Thus, $N = 0.72x = 0.72(400) = \boxed{288}.$ We can also do $1.28y = 1.28(225) = 288.$