jmc

Let the first odd integer be $2n+1$, $n\ge 0$. Then the final odd integer is $2n+1+2(j-1)=2(n+j)-1$. The odd integers form an arithmetic sequence with sum $N=j\left(\frac{(2n+1)+(2(n+j)-1)}{2}\right)=j(2n+j)$. Thus, $j$ is a factor of $N$. Since $n\ge 0$, it follows that $2n+j\ge j$ and $j\le \sqrt{N}$. Since there are exactly $5$ values of $j$ that satisfy the equation, there must be either $9$ or $10$ factors of $N$. This means $N=p_1^2{p}_2^2$ or $N=p_1{p}_2^4$. Unfortunately, we cannot simply observe prime factorizations of $N$ because the factor $(2n+j)$ does not cover all integers for any given value of $j$. Instead we do some casework: If $N$ is odd, then $j$ must also be odd. For every odd value of $j$, $2n+j$ is also odd, making this case valid for all odd $j$. Looking at the forms above and the bound of $1000$, $N$ must be $$(3^2\cdot 5^2),(3^2\cdot 7^2),(3^4\cdot 5),(3^4\cdot 7),(3^4\cdot 11)$$ Those give $5$ possibilities for odd $N$. If $N$ is even, then $j$ must also be even. Substituting $j=2k$, we get $$N=4k(n+k)⟹\frac{N}{4}=k(n+k)$$ Now we can just look at all the prime factorizations since $(n+k)$ cover the integers for any $k$. Note that our upper bound is now $250$: $$\frac{N}{4}=(2^2\cdot 3^2),(2^2\cdot 5^2),(2^2\cdot 7^2),(3^2\cdot 5^2),(2^4\cdot 3),(2^4\cdot 5),(2^4\cdot 7),(2^4\cdot 11),(2^4\cdot 13),(3^4\cdot 2)$$ Those give $10$ possibilities for even $N$. The total number of integers $N$ is $5+10=\boxed{15}$.