12. 3906.

Let $b_{0}=0, b_{1}=a_{1}, b_{2}=a_{2}+a_{1}, b_{119}=\sum_{\mathrm{T}=1}^{119} a_{1}$, here $1 \leqslant b_{1}119$, $b_{1} \neq 119$. There exists $1 \leqslant m<n \leqslant 119, b_{n}-b_{m}=0(\bmod 119)$, i.e., $b_{n}-b_{m}=119 t, t \in \mathbf{N}^{*}, \sum_{1=m \mid 1}^{n} a_{1} = 1\lfloor 9 t$. From the problem, we know $t \neq 1$, so $t \geqslant 2$. Therefore, $\sum_{i=m=1}^{n} a_{1} \geqslant 119 \times 2=238$, thus $\sum_{i=1}^{119} a_{i} \geqslant \sum_{i=m=1}^{n} a_{0} \geqslant 238$. For every 119 numbers as a group, their sum is at least 238, i.e., $\sum_{i=1}^{1311} a_{i} \geqslant 238, \cdots, \sum_{i=1786}^{1994} a_{0} \geqslant 238, \sum_{i=1905}^{2002} a_{i} \geqslant 98$. Therefore, $\sum_{i=1}^{203} a_{i} \geqslant 238 \times 16+98=3906$, at this time $a_{114}=a_{298}=\cdots=a_{1594}=120$, and take $a_{3}=1$.
$\therefore$ The sequence $\left\{a_{n}\right\}$ has exactly 2002 terms, each of which is an integer, and satisfies: $\left.\left.(1)\right\} a_{n}\right\}$ no term equals 119;
(2) $\mid a_{n}$ the sum of any consecutive several terms does not equal 119. Let $S=a_{1}-a_{2}+\cdots+a_{2002}$, then the minimum value of $S$ is $\qquad$.