1. Let the desired number be $\overline{a b c d}=1000 a+100 b+10 c+d$. Then the other number will be $\overline{d c b a}=1000 d+100 c+10 b+a$. The sum of these numbers will be $1001 a+110 b+110 c+1001 d$. Since the division of this sum by 143 resulted in 136, we have the equation: $1001(a+d)+110(b+c)=19448$. Since $(b+c)$ does not exceed 18, and thus $110(b+c)$ does not exceed 1980, we can deduce from the equation that the equality is only possible when $a=d=9$. Then we have: $110(b+c)=19448-18018$, which is equivalent to the equation: $b+c=13$. By enumeration, we find all possible values for the letters $b$ and $c$, and thus obtain 6 variants of four-digit numbers that satisfy the condition: 9949, 9859, 9769, 9679, 9589, 9499.
Answer: 9949, 9859, 9769, 9679, 9589, 9499.