解：解不等式?$x + 2k\geqslant 3k + 4，$?得?$x\geqslant k + 4$?
?$ $?解不等式?$3x + 1\leqslant k，$?得?$x\leqslant \frac {k - 1}3$?
∵不等式組無(wú)解，∴?$k + 4＞\frac {k - 1}3$?
解得?$k＞-\frac {13}2$?
?$ $?解方程?$kx - 2(x - 2)+3x + 6 = 0$?得?$x=\frac {-10}{k + 1}$?
∵方程有正整數(shù)解，∴?$\frac {-10}{k + 1}＞0，$?且?$\frac {-10}{k + 1}$?是整數(shù)
又∵?$k＞-\frac {13}2$?
?$ $?當(dāng)?$\frac {-10}{k + 1}=1$?時(shí)，?$k=-11($?不滿足?$k＞-\frac {13}2，$?舍去)
?$ $?當(dāng)?$\frac {-10}{k + 1}=2$?時(shí)，?$k=-6$?
?$ $?當(dāng)?$\frac {-10}{k + 1}=5$?時(shí)，?$k=-3$?
?$ $?當(dāng)?$\frac {-10}{k + 1}=10$?時(shí)，?$k=-2$?
∴符合條件的所有整數(shù)?$k$?為?$-6，$??$-3，$??$-2$?
