3. Given 20 weights with integer gram masses can measure items of $1, 2, \ldots, 2017$ grams, with weights only placed on one side of the balance, the minimum value of the maximum weight mass is $\qquad$ grams.
3. 147

Analysis: On one hand, let the masses of these 20 weights be $x_{1} \leq x_{2} \leq \ldots \leq x_{20}$. To weigh an item of 1 gram, it is known that $x_{1}=1$. Meanwhile, for any integer $k \geq 2$, if $x_{1}+x_{2}+\ldots+x_{k-1}<2017$, then $x_{k} \leq x_{1}+x_{2}+\ldots+x_{k-1}+1$, otherwise, an item of weight $x_{1}+x_{2}+\ldots+x_{k-1}+1$ cannot be weighed. From the above conclusion, we can derive: $x_{1}=1, x_{2} \leq 2, x_{3} \leq 4, x_{4} \leq 8, x_{5} \leq 16, x_{6} \leq 32, x_{7} \leq 64, x_{8} \leq 128$, and $x_{9}+x_{10}+\ldots+x_{20} \geq 1762$, so $x_{20} \geq \frac{1762}{12}=146 \frac{5}{6}$, which means $x_{20} \geq 147$.

On the other hand, 1 gram, 2 grams, 4 grams, 8 grams, 16 grams, 32 grams, 64 grams, and 128 grams weights, each 1, and 147 grams weights, each 12, can weigh items of $1,2, \mathrm{~L}, 2017$ grams.