Solution:

Since $7 = 2 + 5 = 3 + 4,$ one of $\tau(n), \tau(n+1)$ equals $3$ or $5,$ and $\tau = 3$ means a prime square $p^2$ while $\tau = 5$ means a prime fourth power $p^4.$ So one of $n, n+1$ lies in $\{4, 9, 25, 49, 121, 169, 289, 361, \ldots\} \cup \{16, 81, 625, \ldots\},$ and its neighbor must have $\tau = 4$ (for a square) or be prime (for a fourth power).

Checking neighbors in increasing order: $n = 8$ works $(\tau(8) = 4,$ $\tau(9) = 3);$ $n = 9$ works $(\tau(10) = 4);$ $n = 16$ works $(\tau(16) = 5,$ $17$ prime$);$ $n = 25$ works $(\tau(26) = 4).$ Then $49, 81, 169,$ and $289$ all fail: $\tau(48) = 10,$ $\tau(50) = 6,$ $\tau(80) = 10,$ $82$ is not prime, $\tau(168) = 16,$ $\tau(170) = 8,$ $\tau(288) = 18,$ $\tau(290) = 8.$ Next, $n = 121$ works $(\tau(122) = 4)$ and $n = 361$ works $(\tau(362) = 4).$

The six least solutions are $8, 9, 16, 25, 121, 361,$ with sum $540.$