2¹² = 4096 cells of memory could be used.

The only way for instructions to use a memory cell as data is by referencing it as the operand in an instruction such as STO, LOD, or ADD. Based on the existing instruction formats, the opcode is 4 bits and the operand is a 12-bit reference to a memory cell containing the operand. Similarly, all of the jump instructions have 12-bit operands, resetting the PC to a 12-bit memory address. Given that 12 bits are being used to address a memory cell, there can be at most 2¹² distinct such addresses.

21 was the input.

Recall instructions are numbered from 0, not 1. The only way for instruction 3 to be skipped is if the accumulator held value 0 at the time instruction 2 (JZR BYE) was executed. The value inthe accumulator at that time came from the subtraction in step 1, subtracting VAL (21) from the input. Thus, the input must have been 21 for that subtraction to result in 0.

0011000000001101
0111000000000000
1111000000000000

Please note that the problem asked for machine code as a solution. Of course, the above is equivalent to the assembly code,
LOD 13
OUT
STO

There is more than one way to succeed. Here is one possible solution:

     INP
     STO A
     INP
     STO B
     LOD A
     SUB B
     OUT
     STP
A    DAT
B    DAT
 

There is more than one way to succeed. Here is one possible solution:

     INP
     STO A
     INP
     STO B
     SUB A
     JNG MAXA    ; A is the bigger
     LOD B       ; else B is the bigger
     JMP DONE    
MAXA LOD A
DONE OUT         ; output the answer
     STP
A    DAT
B    DAT
 

     INP
     STO D
     INP         ; this is N
LOOP SUB D       ; subtract D (again)
     JNG NEG     ; until we get negative number
     JMP LOOP
NEG  ADD D       ; we subtracted one time too many
     OUT
     STP
D    DAT
 

     INP
LOOP STO CNT     ; we'll need this
     ADD SUM     ; add current cnt to sum
     STO SUM     ; and re-store new sum
     LOD CNT
     SUB ONE     ; subtract 1 from count
     JZR DONE    ; exit loop if count reaches zero
     JMP LOOP
DONE LOD SUM     ; reload the answer
     OUT
     STP
SUM  DAT 0
CNT  DAT
ONE  DAT 1