Paillier Encryption: problems with double encryption
No, this is not possible. The homomorphism only works at one layer.

A ciphertext in Paillier is $g^m\cdot r^n\bmod{n^2}$. The plaintext space is the multiplicative group of integers modulo $n$. So, for it to even have a chance to work, first of all, the modulus of the outer encryption would have to be greater than $n^2$, where $n$ is the modulus of the inner. Assume this is true. Let the moduli be $n_i$ and $n_o$ for the inner and outer encryptions respectively (and $n_o &gt; n_i^2$ (I'll similarly subscript all the other parameters).

So, for the encryption of $m_1$ we get $g_o^{g_i^{m_1}\cdot r_i^{n_i}\bmod{n_i^2}}\cdot r_o^{n_o}\bmod{n_o^2}$. Now, what we want is something like $g_o^{g_i^{m_1+m_2}\cdot r_i^{n_i}\bmod{n_i^2}}\cdot r_o^{n_o}\bmod{n_o^2}$. Possible options to get there include exponentiating the first by $m_2$, but that doesn't work. We could multiply the first by encryption of $m_2$ with the outer key, but that doesn't work. Even multiplying by the double encryption (first with inner, next with outer) doesn't work.

I know this isn't a rigorous proof, but there doesn't seem to be a way. The algebraic structure is just too mangled after a second encryption for it to work.