方程$4x+y=11$，
解得$y=11-4x$，
当$x=1$时，$y=7$；当$x=2$时，$y=3$，
则方程的正整数解为$\left\{\begin{array}{l}{x=1}\\{y=7}\end{array}\right.$，$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$.
故答案为：$\left\{\begin{array}{l}{x=1}\\{y=7}\end{array}\right.$，$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$.

方程$4x+y=11$，
解得$y=11-4x$，
当$x=1$时，$y=7$；当$x=2$时，$y=3$，
则方程的正整数解为$\left\{\begin{array}{l}{x=1}\\{y=7}\end{array}\right.$，$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$.
故答案为：$\left\{\begin{array}{l}{x=1}\\{y=7}\end{array}\right.$，$\left\{\begin{array}{l}{x=2}\\{y=3}\end{array}\right.$.