6. Let $m$ be a given positive integer, then the largest positive integer $n$ such that $m^{2}+n$ divides $n^{2}+m$ is

保留了原文的换行和格式。
6. $m^{4}-m^{2}+m$. From the problem, we know $\frac{n^{2}+m}{n+m^{2}} \in \mathbf{Z}$. Since $\frac{n^{2}+m}{n+m^{2}}=\frac{\left(n+m^{2}\right)^{2}-2 m^{2}\left(n+m^{2}\right)+m^{4}+m}{n+m^{2}}=$ $n+m^{2}-2 m^{2}+\frac{m^{4}+m}{n+m^{2}}$, therefore, $\frac{m^{4}+m}{n+m^{2}}$ is a positive integer. Thus, $n+m^{2} \leqslant m^{4}+m, n \leqslant m^{4}-m^{2}+m$. Hence, the maximum value of $n$ is $m^{4}-m^{2}+m$.