1. Let the set be
$\left\{\left.\frac{3}{a}+b \right\rvert\, 1 \leqslant a \leqslant b \leqslant 4\right\}$, where the maximum and minimum elements are $M, N$, respectively. Then $M-N=$ $\qquad$ .
Fix $a$, consider the function $y=\frac{3}{a}+x, x \in[a, 4]$ is monotonically increasing, then $y \in\left[\frac{3}{a}+a, \frac{3}{a}+4\right], a \in[1,4]$. Thus $M=\max _{a \in[1,4]}\left\{\frac{3}{a}+4\right\}=7 ; N=\min _{a \in[1,4]}\left\{\frac{3}{a}+a\right\}=2 \sqrt{3}$.
Therefore, $M-N=7-2 \sqrt{3}$.