24. It is given that $m$ and $n$ are two positive integers such that
$$
n-\frac{m}{n}=\frac{2011}{3} .
$$

Determine the smallest possible value of $m$.
24. Answer: 1120
Let $m$ and $n$ be positive integers satisfying the given equation. Then $3\left(n^{2}-m\right)=$ 2011n. Since 2011 is a prime, 3 divides $n$. By letting $n=3 k$, we have $(3 k)^{2}=$ $m+2011 k$. This implies that $k$ divides $m$. Let $m=r k$. Then $9 k^{2}=r k+2011 k$ so that $9 k=r+2011$. The smallest positive integer $r$ such that $r+2011$ is divisible by 9 is $r=5$. Thus $k=(5+2011) / 9=224$. The corresponding values of $m$ and $n$ are $m=1120$ and $n=672$.