13. If $n^{2}+100$ can be divided by $n+10$, then, the maximum positive integer value of $n$ that satisfies the condition is $\qquad$
13. 190.

Since $n^{2}+100$ and $n^{2}-100$ are both divisible by $n+10$, therefore, 200 is divisible by $n+10$.
Hence, the largest positive integer that satisfies the condition is 190.