If \(m\) and \(n\) are two positive integers such that \(7m + 11n = 200\), then the minimum possible value of \(m + n\) is

The minimum of $m+n$ is $20$. Reducing $7m+11n=200$ modulo $7$ gives $n\equiv1\pmod7$, and positivity restricts $n$ to $1,8,15$. The corresponding sums are $28,24,20$, so stopping at $n=8$ gives the higher sum $24$.