3. If $n$ is a positive integer, and $n^{2}+9 n+98$ is exactly equal to the product of two consecutive positive integers, then all values of $n$ are $\qquad$
$3.34,14,7$.
It is easy to know that for any positive integer $n$, the following inequality always holds:
$$
n^{2}+9 n+20<n^{2}+9 n+98<n^{2}+19 n+90 \text {, }
$$

i.e., $(n+4)(n+5)<n^{2}+9 n+98<(n+9)(n+10)$.
$\therefore n^{2}+9 n+98=(n+5)(n+6)$,
or $n^{2}+9 n+98=(n+6)(n+7)$,
or $n^{2}+9 n+98=(n+7)(n+8)$,
or $n^{2}+9 n+98=(n+8)(n+9)$.
From (1) we get $n=34$; from (2) we get $n=14$; from (3) we get $n=7$; (4) has no positive integer solution.
$$
\therefore n=34,14,7 \text {. }
$$