First rule out $n = 1;$ then the split must be $\{2, 5\}$ or $\{3, 4\},$ and values $3$ and $5$ force a prime square $p^2$ or fourth power $p^4$

So one of $n, n+1$ is among $4,$ $9,$ $25,$ $49,$ $121,$ $169,$ $289,$ $361,$ $\ldots$ or $16,$ $81,$ $625,$ $\ldots;$ test each neighbor for $\tau = 4$ or primality

The case $n = 1$ gives $\tau(1) + \tau(2) = 3,$ so any solution has $n \ge 2$ and both divisor counts are at least $2.$ Thus $7 = 2 + 5 = 3 + 4,$ so one of $\tau(n), \tau(n+1)$ equals $3$ or $5.$ Now $\tau = 3$ 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 $540.$