You can find a very, very similar answer here How bizarre that one deals with the maximum and the other deals with the minimum...
mov di,1000h
Because in your 2 hexdigits input routine the biggest value that the user can type is "FF" (255), you could reserve a buffer of that size in your program's data section instead of just trusting that the offset address 1000h will not overlap with anything important in memory.
Data Segment
buf db 256 dup (0)
msg db 0dh,0ah,"Please enter the length of the array: $"
...
To tackle the task of getting the index of the array element where the maximum is located, you can save the current value of the address in DI in an extra register like SI and update that each time the code stumbles upon a bigger value:
...
mov di, offset buf
mov al, [di]
mov max, al
mov si, di ; Address of the first chosen maximum
chk:
mov al, [di]
cmp al, max
jbe a
mov max, al ; Bigger than anything before
mov si, di ; Remember the address of the newest maximum
a:
inc di
loop chk
After this code the index that you're looking for is obtained from subtracting the start of the array from the address of where the maximum was found:
sub si, offset buf ; The index of the maximum
With introducing that SI address comes the opportunity of no longer needing the separate max variable:
...
mov di, offset buf
mov si, di ; Address of the first chosen maximum
chk:
mov al, [di]
cmp al, [si] ; At [si] is the current maximum
jbe a
mov si, di ; Remember the address of the newest maximum
a:
inc di
dec cx
jnz chk
;;; mov al, [si] ; The maximum should you need it
sub si, offset buf ; The index of the maximum
Please note that I've removed the slower loop chk instruction and replaced it by the pair dec cx jnz chk.