jmc

We have $P(n_1)=n_1+3$. Using the property that $a-b\mid P(a)-P(b)$ whenever $a$ and $b$ are distinct integers, we get $$n_1-17\mid P(n_1)-P(17)=(n_1+3)-10=n_1-7,$$and $$n_1-24\mid P(n_1)-P(24)=(n_1+3)-17=n_1-14.$$Since $n_1-7=10+(n_1-17)$ and $n_1-14=10+(n_1-24)$, we must have $$n_1-17\mid 10\text{and}{n}_1-24\mid 10.$$We look for two divisors of $10$ that differ by $7$; we find that $\{2,-5\}$ and $\{5,-2\}$ satisfies these conditions. Therefore, either $n_1-24=-5$, giving $n_1=19$, or $n_1-24=-2$, giving $n_1=22$. From this, we conclude that $n_1,n_2=\boxed{19,22}$.