解：

assume cs:code

stack segment
	db 128 dup(0)
stack ends

code segment
start:
	mov ax,stack
	mov ss,ax
	mov sp,128
	
	push cs
	pop ds ;令ds = cs
	
	mov ax,0
	mov es,ax ;令es = 0
	
	mov si,offset int9 
	mov di, 204h
	mov cx,offset int9end - offset int9
	cld
	
	rep movsb ;把int9安装到了 0:204 
	
	push es:[9*4]
	pop es:[200h]
	push es:[9*4+2]
	pop es:[202h] ;把原本int9的入口保存在了0:200,0:202
	
	cli
	mov word ptr es:[9*4],204h 
	mov word ptr es:[9*4+2],0 ;设置9号中断的入口为0:204
	sti 
	
	mov ax,4c00h
	int 21h
	
int9:
	push ax
	push bx
	push cx
	push es
	
	in al,60h ; 从60h端口读入
	
	mov bx,0
	mov es,bx
	
	pushf
	call dword ptr es:[200h] ;在0:200的位置读取CS,IP，运行原本的int9 
	
	cmp al,9EH
	jne int9ret
	
	mov ax,0b800h
	mov es,ax
	mov bx,0
	mov cx,2000
s:
	mov byte ptr es:[bx],41H
	add bx,2
	loop s
	
int9ret:
	pop es
	pop cx
	pop bx
	pop ax
	
	iret
int9end:
	nop
	
code ends
end start

实验结果：