Answer: 15

Small Hint:

A sum of consecutive odd numbers is a difference of squares: it equals $m^2 - k^2 = (m - k)(m + k).$

Big Hint:

So count factorizations $N = ab$ with $a \le b$ of equal parity: exactly five occur when $N$ (odd case) or $\frac{N}{4}$ (even case) has $9$ or $10$ divisors.

Solution:

The sum of the $(k+1)$th through $m$th positive odd integers is $m^2 - k^2 = (m - k)(m + k).$ Writing $a = m - k$ and $b = m + k,$ the representations of $N$ correspond exactly to the factorizations $N = ab$ with $a \le b$ and $a, b$ of the same parity (then $m = \frac{a + b}{2},$ $k = \frac{b - a}{2}$). So we need $N$ to have exactly $5$ such factorizations.

If $N$ is odd, every divisor pair works, so $N$ needs $9$ or $10$ divisors, i.e. $N = p^8,$ $p^9,$ $p^2q^2,$ or $pq^4$ with $p, q$ distinct odd primes. Below $1000,$ $p^8$ and $p^9$ are impossible, $p^2 q^2$ gives $225$ and $441,$ and $pq^4$ gives $3^4 \cdot 5,$ $3^4 \cdot 7,$ $3^4 \cdot 11:$ five odd values.

If $N$ is even, both factors must be even, so $N = 4M$ and the factorizations correspond to divisor pairs of $M,$ with no parity restriction; we need $M \lt 250$ with $9$ or $10$ divisors. With $9$ divisors: $36,$ $100,$ $196,$ $225.$ With $10$ divisors ($pq^4$): $3 \cdot 2^4,$ $5 \cdot 2^4,$ $7 \cdot 2^4,$ $11 \cdot 2^4,$ $13 \cdot 2^4,$ $2 \cdot 3^4.$ That is $4 + 6 = 10$ even values, for a total of $5 + 10 = 15.$