jmc

In order to obtain a sum of $7$, we must have: either a number with $5$ divisors (a fourth power of a prime) and a number with $2$ divisors (a prime), or a number with $4$ divisors (a semiprime or a cube of a prime) and a number with $3$ divisors (a square of a prime). (No integer greater than $1$ can have fewer than $2$ divisors.) Since both of these cases contain a number with an odd number of divisors, that number must be an even power of a prime. These can come in the form of a square-like $3^2$ with $3$ divisors, or a fourth power like $2^4$ with $5$ divisors. We then find the smallest such values by hand. $2^2$ has two possibilities: $3$ and $4$ or $4$ and $5$. Neither works. $3^2$ has two possibilities: $8$ and $9$ or $9$ and $10$. $(8,9)$ and $(9,10)$ both work. $2^4$ has two possibilities: $15$ and $16$ or $16$ and $17$. Only $(16,17)$ works. $5^2$ has two possibilities: $24$ and $25$ or $25$ and $26$. Only $(25,26)$ works. $7^2$ has two possibilities: $48$ and $49$ or $49$ and $50$. Neither works. $3^4$ has two possibilities: $80$ and $81$ or $81$ and $82$. Neither works. $11^2$ has two possibilities: $120$ and $121$ or $121$ and $122$. Only $(121,122)$ works. $13^2$ has two possibilities: $168$ and $169$ or $169$ and $170$. Neither works. $17^2$ has two possibilities: $288$ and $289$ or $289$ and $290$. Neither works. $19^2$ has two possibilities: $360$ and $361$ or $361$ and $362$. Only $(361,362)$ works. Having computed the working possibilities, we take the sum of the corresponding values of $n$: $8+9+16+25+121+361=\boxed{540}$.