10. The smallest positive integer that can be expressed as the sum of 9 consecutive integers, the sum of 10 consecutive integers, and the sum of 11 consecutive integers is $\qquad$ .
10. 495 .
$$
\begin{array}{l}
\text { Let } t=l+(l+1)+\cdots+(l+8) \\
=m+(m+1)+\cdots+(m+9) \\
=n+(n+1)+\cdots+(n+10)\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$

Then $l=n+2+\frac{2 n+1}{9}$,
$$
m=\frac{n}{10}+n+1 .
$$

Therefore, $2 n+1 \equiv 0(\bmod 9)$,
$$
n \equiv 0(\bmod 10) \text {. }
$$

Thus, the smallest positive integer $n$ that satisfies the condition is $n=40$, i.e.,
$$
t_{\min }=40+41+\cdots+50=495 .
$$