Solution:

We will show that any number of the form $n=2^{p-1} m$ where $m$ is a positive integer that has exactly $k-1$ prime factors all of which are greater than $3$ and $p$ is a prime number such that $(5 / 4)^{(p-1) / 2}>m$ satisfies the given condition.

Suppose that $a$ and $b$ are positive integers such that $a+b=n$ and $d(n) \mid d\left(a^{2}+b^{2}\right)$. Then $p \mid d\left(a^{2}+b^{2}\right)$. Hence $a^{2}+b^{2}=q^{c p-1} r$ where $q$ is a prime, $c$ is a positive integer and $r$ is a positive integer not divisible by $q$. If $q \geq 5$, then
$$2^{2 p-2} m^{2}=n^{2}=(a+b)^{2}>a^{2}+b^{2}=q^{c p-1} r \geq q^{p-1} \geq 5^{p-1}$$
gives a contradiction. So $q$ is $2$ or $3$.

If $q=3$, then $a^{2}+b^{2}$ is divisible by $3$ and this implies that both $a$ and $b$ are divisible by $3$. This means $n=a+b$ is divisible by $3$, a contradiction. Hence $q=2$.

Now we have $a+b=2^{p-1} m$ and $a^{2}+b^{2}=2^{c p-1} r$. If the highest powers of $2$ dividing $a$ and $b$ are different, then $a+b=2^{p-1} m$ implies that the smaller one must be $2^{p-1}$ and this makes $2^{2 p-2}$ the highest power of $2$ dividing $a^{2}+b^{2}=2^{c p-1} r$, or equivalently, $c p-1=2 p-2$, which is not possible. Therefore $a=2^{t} a_{0}$ and $b=2^{t} b_{0}$ for some positive integer $t<p-1$ and odd integers $a_{0}$ and $b_{0}$. Then $a_{0}^{2}+b_{0}^{2}=2^{c p-1-2 t} r$. The left side of this equality is congruent to $2$ modulo $4$, therefore $c p-1-2 t$ must be $1$. But then $t<p-1$ gives $(c / 2) p=t+1<p$, which is not possible either.