The solution below may be helpful to Melanie of West Flegg Middle School, who sent in a very well reasoned, part solution to the problem.

Jack, Paul and Matthew also sent in a correct solution, and again, the reasoning below may help to explain why it works.

Like the number $15$, all odd numbers can be written as the sum of two consecutive integers.

For any odd number $k$, we know that $k-1$ and $k+1$ are even, also $(k-1)/2$ and$(k+1)/2$ are consecutive integers, so we can write $k=(k-1)/2+(k+1)/2$.

All numbers of the form $N = pk$ where $p$ and $k$ are integers, and $k$ is odd and greater than one, can be written as the sum of consecutive integers.

As before we use the integers $(k -1)/2$ and $(k +1)/2$ but this time we have to add $2p$ consecutive integers, $p$ of them less than $(k /2)$ and $p$ of them greater than $(k /2)$, so that their mean is $k /2$ and their sum is $(2 p ) \times( k /2) = pk$

For example $N = 44 = 4 \times 11$ can be written as the sum of $8$ consecutive integers, $4$ less than $5.5$ and $4$ greater than $5.5$ so that the sum is $8 \times5.5 = 44$.

\begin{eqnarray}44 &=& (11 - 7)/2 + (11 - 5)/2 + (11 - 3)/2 + (11 - 1)/2 + (11 + 1)/2 + (11 + 3)/2 + (11 + 5)/2 + (11 + 7)/2 \\ &=& 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9$$ end{eqnarray}

This construction works for all numbers of the form $N = pk$ where $p$ and $k$ are integers and $k$ is odd and greater than one, but the consecutive sum may include some negative integers. If this happens then a string of terms reduces to zero giving fewer terms in the final expression.

By this method,

\begin{eqnarray}12 &=& 4 \times3 \\ &=& (-2) + (-1) + 0 + 1 + 2 + 3 + 4 + 5 \\ &=& 3 + 4 + 5 \end{eqnarray}

giving three terms in the final expression rather than eight.

If a number is a power of $2$ then it is impossible to write it as a sum of consecutive integers. Consider the sum of consecutive integers starting at $( q + 1)$ and ending at $p$ . There are $( p - q )$ terms in this sum. The mean of the terms is $( p + q + 1)/2$ so we can write:

$$( q + 1) + ( q + 2) + \ldots+( p - 1) + p = ( p - q )( p + q + 1)/2$$

As $( p - q )$ and $( p + q + 1)$ cannot both be even, this means that the sum of consecutive integers must represent a number which has at least one odd factor (other than the factor $1$) so a power of $2$ cannot be written as a sum of consecutive integers.

We use the fact here that if $(p - q)$ is odd then $(p + q)$ is odd and $(p + q + 1)$ is even,

or if $(p - q)$ is even then $(p + q)$ is even and $(p + q + 1)$ is odd.