China National Team Selection Test

If $k$ has a factor of square number greater than $1$, let $t^2\mid k$, $t>1$, then taking $m=n=k-\frac{k}{t}+1$, we see that such $k$ has the properties.

If $k$ has no factor of square number, if there are two primes $p_1,p_2$ such that $(p_1-2)(p_2-2)\ge 4$ and $p_1{p}_2\mid k$. Let $k=p_1{p}_2\cdots p_r$ and $p_1,p_2,\dots ,p_r$ be pairwise different, $r\ge 2$. Since there is at least one of $(p_1-1)p_2{p}_3\cdots p_r+1$ and $(p_1-2)p_2{p}_3\cdots p_r+1$ is coprime with $p_1$ (otherwise $p_1$ divides their difference $p_2{p}_3\cdots p_r$, which is a contradiction), taking this number as the number $m$, then, $1<m<k$, $(m,k)=1$. Similarly, we can take a number $n$ of $(p_2-1)p_1{p}_3\cdots p_r+1$ or $(p_2-2)p_1{p}_3\cdots p_r+1$ such that $1<n<k$, $(n,k)=1$. So $p_1{p}_2\cdots p_r\mid (m-1)(n-1)$, and $$\begin{gathered} m+n\ge (p_1-2)p_2{p}_3\cdots p_r+1+(p_2-2)p_1{p}_3\cdots p_r+1 \\ =k+((p_1-2)(p_2-2)-4)p_3\cdots p_r+2>k. \end{gathered}$$ Such $m,n$ will satisfy the conditions.

If there are no two primes $p_1,p_2$ such that $(p_1-2)(p_2-2)\ge 4$, $p_1{p}_2\mid k$, then it is easy to verify such integer $k\ge 3$ can only be $15$, $30$ or as $p$, $2p$ (where $p$ is an odd prime). It is easy to see that, if $k=p,2p,30$, then there are no $m,n$ satisfying the conditions; if $k=15$, then $m=11,n=13$ satisfy the conditions.

Summing up, integer $k\ge 3$ satisfies the conditions if and only if $k$ is not an odd prime, nor double of an odd prime and nor $30$.