Richard Williamson proved that all odd numbers are polite:
This is because any odd number can be expressed as an even number
+ 1 and hence is the sum of half that even number, and the next
integer. For instance, 13 = 6 + 7, 45 = 22 + 23, etc, where 6 and
22 are half of 12 and 44 respectively. As an odd number can
always be expressed as the sum of two consecutive integers, and
as this satisfies the definition of politeness, all odd numbers
are polite.
Richard also found the consecutive sums for 544 and 434 and the
way of characterizing impolite numbers, namely that they are all
powers of 2. All other integers are polite. Shu Cao of the Oxford
High School for Girls also proved this result.
James of Hills Road Sixth Form College, Cambridge proved this
result and also gave a nice generalization defining a 'degree of
politeness' of numbers.
The desired sequence of integers has sum $0.5n(2a+n-1)$ where $n$
is the number of terms, $a$ is the initial value. If we are
looking for a sequence of integers that sum to 544, $n(2a+n-1) =
544 \times 2 = 1088 = 2^6 \times 17$. A sensible value for n
would therefore be 17. This means that $a =24$, so $544 = 24 + 25
+ 26 + ... + 39 + 40$.
Using the same procedure on 424 yields $424 = 19 + 20 + 21 + ...
+ 34$.
Hypothesis: all numbers of the form $2^m$ (where $m$ is a
positive integer) are impolite.
Proof: The desired sequence of integers which add up to give a
polite number has the sum $0.5n(2a+n-1)$ where $n\geq 2$ and one
of the factors $n$ or $(2a+n-1)$ must be odd and the other even.
Thus every polite number has an odd factor greater than or equal
to 3. If the number is a power of 2 then it does not have an odd
factor so it is impolite.
Hypothesis: all numbers not of the form $2^m$ are polite
If $k$ is a polite number then $2k=n^2 + (2a-1)n$ is equal to
$2^j \times r$ where $r$ is odd (as above). Set whichever is the
smaller of $2^j$ and $r$ equal to $n$. Then either (1) $2^j + 2a
- 1 = r$ if $2^j< r$ or (2) $r + 2a - 1 = 2^j$ if $2^j> r$.
In case (1) $2a = r + 1 - 2^j$. This is greater than 0, as
$2^j< r$, and is also even, as $2^j - 1$ is an odd number and
$r$ is an odd number. (The difference of two odd numbers is
even). There therefore exists an integer value of $a$.
Similarly in case (2) $2a = 2^j + 1 - r$ which is a positive even
number.
This proves that all numbers not of the form $2^m$ are polite,
and all those of the form $2^m$ are impolite.
It is interesting to note that numbers in which $r$ is not prime
may have two or more different sequences of integers which sum to
them. For example
$$\eqalign {140 &= 14+15+16+17+18+19+20+21 \cr
&=26+27+28+29+30 \cr &=17+18+19+20+21+22+23.}$$
An extension of the above problem is to find a general formula
for the 'degree of politeness' of a number. 140 has politeness 3,
as it can be broken down into 3 different sums of consecutive
integers. In general the politeness of a number is $2^k -1$ where
$k$ is the number of prime factors of $r$ if $r$ is less than
$2^j$ in the context used above. Repeated prime factors are
counted repeatedly, for example $3^5$ contributes 5 to the number
of prime factors. The number of different values for n, with 1
excluded, is given by the number of possible divisors of $r$.
The situation is more complicated if some of the factors of $r$
are greater than $2^j$ and some less, but the politeness will
still be equal to the number of different divisors of $r$ which
are less than $2^j$. If $2^j$ is less than all the factors of $r$
then the politeness of the number will be equal to 1.