Q:

Have you been nice this year?

Introduction
Santa has too many names to process, and needs your help! He needs you to write a program or a function, which outputs nice, naughty, very naughty or very very naughty. To determine how nice or naughty someone is, Santa had developed an algorithm:
Nice (division, math):
First of all, we obtain a number from the name by adding all the letters up (spaces ignored). For example:
Doorknob =

D = 4
o = 15
o = 15
r = 18
k = 11
n = 14
o = 15
b = 2

4 + 15 + 15 + 18 + 11 + 14 + 15 + 2 = 94

If the number of divisors is equal to the length of the name, the person is considered nice. This means that your program should output [name] has been nice. Here, the divisors of 94 are:
Divisors of 94: 1, 2, 47, 94

There are 4 divisors, but the name has length 8 (spaces included). Conclusion, Doorknob has not been nice. So we continue our journey:

Naughty (sequence, math):
Santa has developed a new sequence, the christmas number. First, we will look at the following christmas trees:
n = 1        n = 2          n = 3          n = 4

                                             *
                                            ***
                                           *****
                                          *******
                                         *********
                          _   *             ***
                         |   ***           *****
                         |  *****         *******
               *         | *******       *********
  *           ***        |   ***            ***
 ***         *****   n*n |  *****          *****
  *           ***        | *******        *******
             *****       |   ***         *********
               *         |  *****           ***
                         |_*******         *****
                              *           *******
                           |_____|       *********
                             2n+1            *

  5            18             47             98

The amount of asterisks determines the christmas number. The sequence goes as following: 5, 18, 47, 98, 177, ....
From here, we can conclude that 94 is not a christmas number. That means that Doorknob has not just been naughty.

Very naughty (string):
For this, we need to find out if Doorknob is a raising ladder string. This is determined by the letters in the name with A = 1, B = 2, C = 3, etc.:
First we will look at the first letter, the D. This has value 4. This is our starting point. The next letter is o. This has the value 15, which is higher than our previous value, so we're going a step higher on the ladder. The next value is also an o. This is the same, so we're doing nothing. If the next value is higher than the current value, we will go a step higher. If the next value is lower than the current value, we will go a ster lower. If it's the same, we'll stay on the same step. This visualised for Doorknob, Martin Buttner and Alex A:
            O
           / \
      R   N   B
     / \ /
  O-O   K                 T   N   U                 L   X
 /                       / \ / \ / \               / \ / \
D                   M   R   I   B   T-T           A   E   A
                     \ /               \
                      A                 N   R
                                         \ /
                                          E

You can see that Doorknob ended higher than the starting position. So Doorknob has been very naughty. Martin Buttner and Alex A didn't get higher than the starting point. So they are both very very naughty.
Test cases
Input: Doorknob
Output: Doorknob has been very naughty

Input: Martin Buttner
Output: Martin Buttner has been very very naughty

Input: Jakube
Output: Jakube has been nice

Input: B
Output: B has been very very naughty

Rules

You need to provide a program or a function which takes input (which consists of at least one letter).
The input will consist of uppercase letters, lowercase letters and spaces. Spaces are ignored during the process, except for the length of the input.
This is code-golf, so the submission with the least amount of bytes wins!

A:

CJam, 109 108 107 bytes
l" has been "1$[_,\S-:eu'@f-:A:+__,:)f%0e=@=\_{)__2+**))}%&A2ew::-:g1b0<]{}#4,="very "*_5>\"naughty""nice"?

Try it online.
Explanation
A full explanation will have to wait until later, but here is the code broken down into the different sections:
l" has been "1$[                    e# Some preparation...
  _,\S-:eu'@f-:A:+                  e# Determine letter sum.
  __,:)f%0e=@=                      e# Check divisor count.
  \_{)__2+**))}%&                   e# Check for Christmas number.
  A2ew::-:g1b0<                     e# Check ladder.
]{}#4,="very "*_5>\"naughty""nice"? e# Determine nice/naughtiness.

A:

Pyth 86 bytes
Apparently I've been nice this year...
jd[z"has been"?qlzl{yPJsKxL+NG@Grz0"nice"+*"very "&!}Jm+*+2d*dd2SJhgs._M-VKtK0"naughty

Try it online: Demonstration or Test Suite
Explanation:
jd[z"has been"...      list with input string, "has been" and ...; join with spaces
qlzl{yPJsKxL+NG@Grz0   compares the length of input with the number of divisors
                       (computes all prime factors and counts the elements in the powerset)
?..."nice"             if True, use "nice" as last list element
                       else:
}Jm+*+2d*dd2SJ            check for christmas number
                          (checks, if its of the form n^3+2*n^2+2)
gs._M-VKtK0               check if raising ladder string ends lower or equal
                          (assigns each neighbor pair a number -1,0,1 and computes the sum)
&!...h...                 returns 0, 1 or 2
*"very "                  repeat "very " this times
+..."naughty              add "naughty" and use this as the third list element

A:

MATL, 117 bytes
My longest MATL program so far :-) :-(
Uses current release of the language, which is earlier than this challenge.
jttk96-t0>)tstt:\~s4$bn=?xx'nice'}[1,2,0,2]99:ZQ=a?'naughty'}dt0>sw0<s-O>~'very naughty'w?'very 'wh]]]' has been 'whh

Examples
>> matl
 > jttk96-t0>)tstt:\~s4$bn=?xx'nice'}[1,2,0,2]99:ZQ=a?'naughty'}dt0>sw0<s-O>~'very naughty'w?'very 'wh]]]' has been 'whh
 > 
> Doorknob
Doorknob has been very naughty

>> matl
 > jttk96-t0>)tstt:\~s4$bn=?xx'nice'}[1,2,0,2]99:ZQ=a?'naughty'}dt0>sw0<s-O>~'very naughty'w?'very 'wh]]]' has been 'whh
 > 
> Jakube
Jakube has been nice

Explanation
jt                        % input string. Duplicate
tk96-t0>)                 % duplicate. Convert to lower, then to numbers, remove spaces
ts                        % duplicate. Sum
tt:                       % duplicate. Vector from 1 to obtained sum
\~s                       % modulus operation. Count zeros to determine number of divisors
4$bn=                     % does it equal original name length?
?                         % if so
    xx'nice'              % remove values not needed, and push string
}                         % else
    [1,2,0,2]99:ZQ        % all Christmas numbers
    =a                    % does sum equal any Christmas number?
    ?                     % if so
        'naughty'         % push string
    }                     % else
        dt0>s             % total number of increases
        w0<s              % total number of decreases
        -O>~              % subtract both. Is total <=0?
        'very naughty'w   % push string, which will be needed in either case. Swap
        ?                 % if total was <=0
            'very 'wh     % prepend string
        ]                 % end if
    ]                     % end if
]                         % end if
' has been 'whh           % build complete string from pushed parts

