7. 147.

Let the masses of 20 weights be $a_{1} \leqslant a_{2} \leqslant \cdots \leqslant a_{20}$.
We will prove by mathematical induction that: $a_{n} \leqslant 2^{n-1}(n \leqslant 11)$.
When $n=1$, the conclusion is obviously true.
Assume the conclusion holds for $n=1,2, \cdots, k$.
If $a_{k+1}>2^{k}(k \leqslant 10)$, then by
$$
\sum_{i=1}^{k} a_{i} \leqslant \sum_{i=1}^{k} 2^{i-1}=2^{k}-1,
$$

we know that an item of $2^{k}$ grams cannot be weighed, which is a contradiction.
Thus, $\sum_{i=1}^{8} a_{i} \leqslant \sum_{i=0}^{7} 2^{i}=255$.
Hence, $\sum_{i=9}^{20} a_{i} \geqslant 2014-255=1759$.
Therefore, $a_{20} \geqslant \frac{1759}{12}>146 \Rightarrow a_{20} \geqslant 147$.
Moreover, when $a_{k}=2^{k-1}(k \leqslant 8), a_{9}=a_{10}=\cdots=a_{20}$
$=147$, the conditions are satisfied.
Thus, the minimum value is 147.
7. Given 20 weights with integer gram masses, they can measure items of $1,2, \cdots, 2014$ grams, with the weights only placed on one side of the balance. Then the minimum value of the maximum weight mass is $\qquad$ grams.