14.2006 positive integers $a_{1}, a_{2}, \cdots, a_{2006}$, none of which are equal to 119, are arranged in a row, where the sum of any consecutive several terms is not equal to 119. Find the minimum value of $a_{1}+a_{2}+\cdots+$ $a_{2006}$.
14. First prove the proposition:

For any 119 positive integers $b_{1}, b_{2}, \cdots, b_{119}$, there must exist some (at least one, or possibly all) whose sum is a multiple of 119.
In fact, consider the following 119 positive integers:
$$
b_{1}, b_{1}+b_{2}, \cdots, b_{1}+b_{2}+\cdots+b_{119} \text {. }
$$

If one of them is a multiple of 119, then the conclusion holds.
If none of them is a multiple of 119, then the remainders when they are divided by 119 can only be $1,2, \cdots, 118$ in 118 different cases. Therefore, there must be two with the same remainder when divided by 119, say $b_{1}+b_{2}+\cdots+b_{i}$ and $b_{1}+b_{2}+\cdots+b_{j}(1 \leqslant i<j \leqslant 119)$, thus,
$119 \mid \left(b_{i+1}+b_{i+2}+\cdots+b_{j}\right)$.
The proposition is proved.
For any 119 numbers in $a_{1}, a_{2}, \cdots, a_{2006}$, by the proposition, there must be some whose sum is a multiple of 119.

Given that it is not equal to 119, it must be greater than or equal to $2 \times 119$.
Since $2006=16 \times 119+102$, we have:
$$
a_{1}+a_{2}+\cdots+a_{2006} \geqslant 16 \times 238+102=3910 \text {. }
$$

If $a_{119}=a_{288}=\cdots=a_{1904}=120$, and the rest are all 1, then the equality holds.
Therefore, the minimum value of $a_{1}+a_{2}+\cdots+a_{2006}$ is 3910.