解:$ (1)P=\frac {826 - 628}{22}=9,$
因为$9$为整数,
所以$826$为$“$有趣数$”;$
$P=\frac {326 - 623}{22}=-13.5,$
因为$-13.5$不是整数,
所以$326$不是$“$有趣数$”;$
$ (2)$因为$f = 100x + 42,$$s = 120 + y(1\leqslant x\leqslant 9,$
$1\leqslant y\leqslant 9,$且$x,$$y$均为整数$),$$f$和$s $的$“$有趣值$”$
分别记为$P_{1}$和$P_{2},$
所以$P_{1}=\frac {100x + 42-(240 + x)}{22}=\frac {99x - 198}{22}=\frac {9(x - 2)}{2},$
$P_{2}=\frac {120 + y-(100y + 21)}{22}=\frac {99 - 99y}{22}=\frac {9(1 - y)}{2}。$
因为$P_{1}-2P_{2} = 36,$
所以$\frac {9(x - 2)}{2}-2×\frac {9(1 - y)}{2}=36,$
整理可得$x + 2y = 12。$
因为$1\leqslant x\leqslant 9,$$1\leqslant y\leqslant 9,$且$x,$$y$均为整数,
所以$\begin {cases}x = 2 \\y = 5 \end {cases},$$\begin {cases}x = 4 \\y = 4 \end {cases},$$\begin {cases}x = 6 \\y = 3 \end {cases},$$\begin {cases}x = 8\\y = 2 \end {cases}。$
$ $将$\begin {cases}x = 2\\y = 5 \end {cases}$代入,可得
$P_{1}=\frac {9×(2 - 2)}{2}=0,$
$P_{2}=\frac {9×(1 - 5)}{2}=-18,$符合题意,
所以$\begin {cases}f = 242\\s = 125 \end {cases};$
将$\begin {cases}x = 4\\y = 4 \end {cases}$代入,可得
$P_{1}=\frac {9×(4 - 2)}{2}=9,$
$P_{2}=\frac {9×(1 - 4)}{2}=-13.5,$$-13.5$不是整数,不符
合题意;
将$\begin {cases}x = 6\\y = 3 \end {cases}$代入,可得
$P_{1}=\frac {9×(6 - 2)}{2}=18,$
$P_{2}=\frac {9×(1 - 3)}{2}=-9,$符合题意,
所以$\begin {cases}f = 642\\s = 123 \end {cases};$
将$\begin {cases}x = 8\\y = 2 \end {cases}$代入,可得
$P_{1}=\frac {9×(8 - 2)}{2}=27,$
$P_{2}=\frac {9×(1 - 2)}{2}=-4.5,$$-4.5$不是整数,不符合
题意。
所以满足条件的三位数$f$和$s $分别为$\begin {cases}f = 242\\s = 125 \end {cases}$和
$\begin {cases}f = 642\\s = 123 \end {cases}。$