2. If the positive integer $n \geqslant 2006$, and 122 divides $91 n-37$, then the minimum value of $n$ is $\qquad$ .
2.2061 .

Since 122 is even and 91 and 37 are odd, $n$ must be odd.
Let $n=2k-1(k \in \mathbf{Z})$, then
122 | $(91n-37) \Leftrightarrow 122 | [91(2k-1)-37]$
$\Leftrightarrow \frac{182k-128}{122} \in \mathbf{Z} \Leftrightarrow \frac{91k-64}{61} \in \mathbf{Z}$
$\Leftrightarrow \frac{30k-3}{61} \in \mathbf{Z} \Leftrightarrow \frac{60k-6}{61} \in \mathbf{Z} (2$ is coprime with $61)$
$\Leftrightarrow \frac{k+6}{61} \in \mathbf{Z} \Leftrightarrow k+6=61t$ ( $t$ is a positive integer)
$\Leftrightarrow k=61t-6 \Leftrightarrow n=122t-13$.
Let $122t-13 \geqslant 2006$, then $t \geqslant \frac{2019}{122}=16+\frac{67}{122}$, i.e., $t \geqslant 17$.
Thus, the minimum value of $n$ is $122 \times 17-13=2061$.