10. Among the first 100 positive integers, the sum of numbers that can be expressed as the difference of squares of two positive integers is
Since $n=a^{2}-b^{2}=(a+b)(a-b)$, where $a+b$ and $a-b$ have the same parity, when $n$ can be expressed as the difference of squares of two positive integers, $n$ is odd or a multiple of 4.
Conversely, when $n$ is odd or a multiple of 4, decompose $n$ into the product of two odd numbers or two even numbers, making them equal to $a+b$ and $a-b$, respectively. Then, except for 1 and 4, it is always possible to find $a \neq b$ such that $n=a^{2}-b^{2}$.
Since $1+4+2+6+\cdots+98=5+\frac{25 \times 100}{2}=1255$,
the sum of the numbers among the first 100 positive integers that can be written as the difference of squares of two positive integers is $5050-1255=3795$.