Because every list of consecutive integers can be summed to form another integer, but it is undetermined as whether every integer can be described by the sum of consecutive integers, it surely makes sense to consider the sum of various lists rather than to attempt to split a multitude of integers.
Hence, by letting, for some $q$, $x_n$ be the sum of $q$ consecutive integers beginning with $n$;
$$x_n=n+(n+1)+(n+2)+\dots+(n+q-3)+(n+q-2)+(n+q-1)$$
You get the identity;
$$x_n=qn+(1+2+3+\dots+(q-3)+(q-2)+(q-1))=qn+\frac{q(q-1)}{2}$$
From this, it should be easy to see that all odd $q$ result in $x_n$ that are a multiple of $q$, because the denominator of the fraction halves $q-1$, leaving $q$ intact. It should, therefore, be instantly obvious that, because $x_{n+1}-x_n=q$, every multiple of $q$ greater than or equal to $x_1$ can be expressed as the sum of $q$ consecutive integers.
Similarly, for even $q$, it is immediately obvious that $x_n\equiv\frac{q}{2}(\mod q)$. Since $x_{n+1}-x_n=q$ is true regardless of the value of $q$, it is evident that all numbers which leave a remainder of $\frac{q}{2}$ after division by $q$, and which equal or exceed $x_1$, can be written as the sum of $q$ consecutive integers.
By now considering $q=2$, it can be seen that all odd numbers $\geq3$ can be expressed as the sum of 2 consecutive positive integers.
Using the definition of $x_1=\frac{q(q+1)}{2}$, it can be said that, if an integer $a$ has $q$ as a factor and $q$ is odd, there is a requirement that $a\geq\frac{q(q+1)}{2}$ for $a$ to be expressible as the sum of $q$ consecutive integers. By making this a requirement upon $q$, it becomes obvious than any integer $a$ with a factor $q\leq\frac{\sqrt{8a+1}-1}{2}$ can be expressed in this way.
With the same requirement upon $q$, any integers which have $\frac{q}{2}$ as a factor, but not $q$ as a factor, where $q$ is even, can be expressed as the sum of $q$ consecutive integers.
Finally, to work out which numbers can and cannot be expressed as the sum of consecutive integers, consider a number $a=2^p(2b+1)$, with $b$ and $p$ in the natural numbers. Because of this formulation, $a$ can describe any even number except a power of $2$.
Firstly, $a$ may be expressed as $2^{p+1}$ consecutive integers iff (if and only if);
$$2^p(2^{p+1}+1)\leq2^p(2b+1)$$
due to the conditions above. Hence, this then simply becomes the condition that;
$$b\geq2^p$$
Similarly, $a$ may be expressed as $2b-1$ consecutive integers iff;
$$(b+1)(2b+1)\leq2^p(2b+1)$$
and this has the consequence that;
$$b\leq2^p-1$$
Hence, for every possible natural number $b$, there is a way to express the possible corresponding $a$ values as the sum of consecutive integers.
The remaining case is $a=2^p$, with $p$ in the integers, $p\geq0$. The conditions above demonstrate that the only possible number of consecutive integers that might sum to $a$ is $2^{p+1}$, but, as $a<2^p(2^{p+1}+1)$, this isn't possible either. Hence, no power of 2 (including 1) can be expressed as the sum of consecutive integers.
Finally, the number of ways of describing an integer $a$ is determined by the factors of $a$ and their possession of the properties described above.
In summary:
No power of 2 can be written as the sum of consecutive integers.
Any factor of any other number $a$ which is less than $\frac{\sqrt{8a+1}-1}{2}$ generates a way of describing $a$ as the sum of consecutive integers, unless the factor is even and double itself also divides $a$.