$解：(1+n)x+2023y=2m+2的“交換系數(shù)方程”為$
$(2m+2)x+ 2023y=1+n$
$或(1+n)x+(2m+2)y=2023$
$∵(10m-t)x+2023y=m+t是關(guān)于x、y的二元一次方程$
$(1+n)x+2023y=2m+2的“交換系數(shù)方程”$
$∴當(dāng)(10m-t)x+2023y=m+t各系數(shù)與$
$(2m+2)x+2023y=1+n各系數(shù)對應(yīng)相等時$
$得\begin{cases}{10m-t=2m+2}\\{m+t=1+n}\end{cases}，①$
$當(dāng)(10m-t)x+2023y=m+t各系數(shù)與$
$(1+n)x+(2m+2)y=2023各系數(shù)對應(yīng)相等時$
$得\begin{cases}{10m-t=1+n}\\{2023=2m+2}\\{m+t=2023}\end{cases}，②$
$解方程組①得\begin{cases}{m=\frac{t+2}{8}}\\{n=\frac{9t-6}{8}}\end{cases}$
$∵t＜n＜8m$
$∴t＜\frac{9t-6}{8}＜t+2，解得6＜t＜22(t為整數(shù))$
$∴8＜t+2＜24$
$∴若m=\frac{t+2}{8}為整數(shù)，必須有t+2=16$
$此時m=2$
$∴t=14$
$當(dāng)t=14時$
$n=\frac{9t-6}{8}=\frac{9×14-6}{8}=\frac{126-6}{8}=\frac{120}{8}=15$
$符合題意$
$∴m=2$
$解方程組②得$
$m=\frac{2023-2}{2}=\frac{2021}{2}(不是整數(shù))$
$∴方程組②的解不符合題意，需舍去.$
$綜上，m=2.$
