Answer: all numbers $n$ such that $p\mid n$ and $n$ is not a perfect square. Why others don't work: to pair the divisors we obviously need $\tau(n)\equiv0\pmod 2$ which means n is not a perfect square. Now $1$ is involved in a pair $(1,d)$ with $d>1$ so $p\mid d$ but $d\mid n$ so $p\mid n$ and we are done Construction of the pairs for these $n$: We first need a lemma: Lemma:if $x\ge 1$ and $x$ is not a perfect square then it's divisors can be paired in $(d,d')$ such that $d\mid d'$ Let $x=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\dots p_t^{\alpha_t}\cdot p^{\alpha}$ such that WLOG $\alpha$ is odd Then let $s_1,s_2,\dots s_j$ the divisors of $\frac{x}{p^{\alpha}}$ and just pair $(p^{2i}\cdot s_t,p^{2i+1}\cdot s_t)$ for $0\le i\le\frac{ \alpha-1}{2}$ and $1\le t \le j$. Coming back to our problem we have 2 cases: Case 1: $v_p(n)\equiv 1 \pmod 2$ Then using the construction in the lemma all fractions are integers divisible by $p$ so we are done. Case 2: $v_p(n)\equiv 0 \pmod 2$ Let $n=p_1^{\alpha_1}\cdot p_2^{\alpha_2}\dots p_t^{\alpha_t}\cdot p^{\alpha}$ and we know $\alpha$ is even. Then $\frac{x}{p^{\alpha}}$ is not a perfect square and by the lemma let it's divisors be $a_1,\dots a_j,b_1,b_2,\dots b_j$ where $a_i\mid b_i$. So the final divisors are $a_1,a_2,\dots a_j,b_1,b_2,\dots b_j$ $p\cdot a_1,p\cdot a_2,\dots p\cdot a_j,p\cdot b_1,p\cdot b_2,\dots p\cdot b_j$ $p^2 \cdot a_1,p^2 \cdot a_2,\dots p^2 \cdot a_j,p^2 \cdot b_1,p^2 \cdot b_2,\dots p^2 \cdot b_j$ The others lines(in number of $\alpha-2$ which is even) can be paired with the ratio $p$. So it is enough to solve for these $3$ lines. But then just pair $b_1,b_2,\dots b_j$ with $pb_1,pb_2,\dots pb_j$ and $a_1,a_2,\dots a_j$ with $p^2\cdot a_1,p^2\cdot a_2,\dots p^2\cdot a_j$ and in the final $pa_1,pa_2,\dots pa_j$ with $p^2\cdot b_1,p^2\cdot b_2,\dots p^2\cdot b_j$ which works.

I proposed this problem. I hope it was enjoyable for the contest!