Why so odd!

Same way as Patrick. We look for numbers of the form $x = 2^pq^2$ , where $p$ is a non-negative integer and $q$ is odd.

For any given $p$ , the number of values for $q$ is $n_p = \left\lceil\frac{\left\lfloor\sqrt{\frac{5000}{2^p}}\right\rfloor}{2}\right\rceil.$ Thus we have the following: $\begin{array}{r|r} p & n_p \\ \hline 0 & 35 \\ 1 & 25 \\ 2 & 18 \\ 3 & 13 \\ 4 & 9 \\ 5 & 6 \\ 6 & 4 \\ 7 & 3 \\ 8 & 2 \\ 9 & 2 \\ 10 & 1 \\ 11 & 1 \\ 12 & 1 \\ \hline \end{array}$

For example, for $p = 7$ we have the values $x = 2^7\cdot 1^2 = 128; \ \ \ x = 2^7\cdot 3^2 = 1152; \ \ \ x = 2^7\cdot 5^2 = 3200;$ their sums of divisors are 255, 3315, and 7905, respectively.

Adding the numbers in the second column we find a total of 120.

It is well-known that $F$ is multiplicative, i.e. $F(ab) = F(a)F(b)$ if $a$ and $b$ are coprime.

If $a = 2^k$ , $F(a) = 2^{k+1}-1$ is odd. If $b$ is odd, then every divisor of $b$ is odd, so $F(b)$ is odd if and only if the number of divisors of $b$ is odd, which happens if and only if $b$ is a square.

Since every positive integer can be written (uniquely) as a power of $2$ times an odd number, we see that $F(x)$ is odd if and only if $x$ is a power of $2$ times an odd square.

It's trivial (but tedious) to count the number of such $x$ less than or equal to $5000$ . There are $\fbox{120}$ of them.