a)
Clearly the sum of the first two positive integers can be represented as the sum of one positive integer. Now, let $n \ge 2$ and let us show that the sum of the $n + 1$ first positive integers cannot be represented as a sum of $n$ consecutive integers. Indeed, on one hand $1 + 2 + \dots + n + (n+1) > 2 + 3 + \dots + n + (n+1)$, on the other hand $1 + 2 + \dots + n + (n+1) < 1 + 2 + 3 + \dots + n + (n+1) + 1 = 3 + \dots + n + (n+1) + 2 + 2 \le 3 + \dots + n + (n+1) + (n+2)$. So the sum of $n + 1$ first positive integers $1 + \dots + (n+1)$ lies between $2 + \dots + (n+1)$ and $3 + \dots + (n+2)$, which both are consecutive sums of $n$ consecutive integers. So, the number $1 + \dots + (n+1)$ is not a sum of $n$ consecutive integers.

b)
Let $n$ be any positive integer. To solve the problem, it suffices to see that $n^2 + (n^2 + 1) + \dots + (n^2 + n) = n^2 \cdot (n+1) + (1 + \dots + n) = n \cdot (n^2 + n) + (1 + \dots + n) = (n^2 + n + 1) + \dots + (n^2 + n + n)$.

We use the formula for the sum of arithmetic progression.

a) If $n$ is odd, then the sum of $n$ consecutive integers is divisible by $n$. So, if the number $1+2+...+(n+1) = \frac{n+1}{2} \cdot (n+2)$ was the sum of $n$ consecutive integers, it would be divisible by $\frac{n+1}{2}$ and by $n$. As $n+1$ and $n$ are relatively prime, the same clearly holds for $\frac{n+1}{2}$ and $n$. Therefore, $\frac{n+1}{2} \cdot (n+2)$ should be divisible by $\frac{n+1}{2} \cdot n$, meaning that $n+2$ should be divisible by $n$.

If $n$ is even, the sum of $n$ consecutive integers is divisible by $\frac{n}{2}$. So, if $1+2+...+(n+1) = (n+1) \cdot \frac{n+2}{2}$ was the sum of $n$ consecutive integers, it would be divisible by both $n+1$ and $\frac{n}{2}$. As $n+1$ and $n$ are relatively prime, also $n+1$ and $\frac{n}{2}$ are relatively prime. Thus, $(n+1) \cdot \frac{n+2}{2}$ should be divisible by $(n+1) \cdot \frac{n}{2}$, implying that $n+2$ is divisible by $n$.

So, in all cases $n+2$ is divisible by $n$, which is equivalent to saying $2$ is divisible by $n$. So, $n=1$ or $n=2$. Clearly $1+2$ is the sum of one integer, but $1+2+3=6$, being an even number, cannot be represented as the sum of two consecutive integers.

b) If $a$ is the first of the two consecutive integers, then the problem can be represented as the equation $(a+a+n)(n+1)/2 = (a+n+1+a+2n)n/2$. By simplifying we see that it is equivalent to $a = n^2$. This means that the sum of $n+1$ consecutive integers, first of which is $n^2$, is the sum of the next $n$ consecutive integers. So, the desired numbers exist for every $n$.