1. There are weights of $11 \mathrm{~g}$ and $17 \mathrm{~g}$ available in sufficient quantity. To weigh an object of mass $3 \mathrm{~g}$ on a balance, at least $\qquad$ such weights are needed.
Let the weights of 11 g and 17 g be used $x$ times and $y$ times, respectively. Then we have $11 x - 17 y = 3$ or $17 y - 11 x = 3$.
The solutions are $\left\{\begin{array}{l}x=8+17 t, \\ y=5+11 t\end{array}\right.$ or $\left\{\begin{array}{l}x=9+17 t, \\ y=6+11 t .\end{array}\right.$
where $t$ is an integer.
Since $x$ and $y$ are positive integers, the minimum values of $x$ and $y$ can be
$$
\left\{\begin{array}{l}
x=8, \\
y=5
\end{array} \text { or } \left\{\begin{array}{l}
x=9, \\
y=6 .
\end{array}\right.\right.
$$

Therefore, the minimum value of $x+y$ is 13.