Call a positive integer with this property good and let $n$ be an arbitrary positive integer. Since $n$ is good implies any $d | n$ is also good we can assume that $n = 2^{2r} \cdot 3^{2s} \cdot l^2$ where $r, s$ are positive integers and $(l, 6) = 1$. To show that $n$ is good it suffices to show that there exist positive integers $a, b$ so that $n | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$. Choose a positive integer $b$ so that $2^{2r} \cdot l^2 | 3^{2s} \cdot b^2 - 1 \Rightarrow b \equiv 3^{-s} \pmod {2^{2r} \cdot l^2}$. This is well defined since $(2l, 3) = 1$. Similarly, choose $a$ so that $3^{2s} | (2^r \cdot l)^2 \cdot a^2 - 1 \Rightarrow a \equiv (2^r \cdot l)^{-1} \pmod {3^{2s}}$. This is well defined as above. Now we also have $2^{2r} \cdot l^2 | (2^r \cdot l)^2 \cdot a^2$ hence $2^{2r} \cdot l^2 | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$. Similarly, we have $3^{2s} | 3^{2s} \cdot b^2$ hence $3^{2s} | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$. Finally, we have $2^{2r} \cdot 3^{2s} \cdot l^2 | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$ hence $n$ is good as desired.

mathl25 wrote: Call a positive integer with this property good and let $n$ be an arbitrary positive integer. Since $n$ is good implies any $d | n$ is also good we can assume that $n = 2^{2r} \cdot 3^{2s} \cdot l^2$ where $r, s$ are positive integers and $(l, 6) = 1$. To show that $n$ is good it suffices to show that there exist positive integers $a, b$ so that $n | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$. Choose a positive integer $b$ so that $2^{2r} \cdot l^2 | 3^{2s} \cdot b^2 - 1 \Rightarrow b \equiv 3^{-s} \pmod {2^{2r} \cdot l^2}$. This is well defined since $(2l, 3) = 1$. Similarly, choose $a$ so that $3^{2s} | (2^r \cdot l)^2 \cdot a^2 - 1 \Rightarrow a \equiv (2^r \cdot l)^{-1} \pmod {3^{2s}}$. This is well defined as above. Now we also have $2^{2r} \cdot l^2 | (2^r \cdot l)^2 \cdot a^2$ hence $2^{2r} \cdot l^2 | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$. Similarly, we have $3^{2s} | 3^{2s} \cdot b^2$ hence $3^{2s} | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$. Finally, we have $2^{2r} \cdot 3^{2s} \cdot l^2 | (2^r \cdot l)^2 \cdot a^2 + (3^s)^2 \cdot b^2 - 1$ hence $n$ is good as desired. GOOD! but I still want to ask: $$ \mathbf{Do}\,\,\mathbf{other}\,\,\mathbf{solutions}\,\,\mathbf{exist}? $$