Solution:

Dividing by $7$ turns the summands into $1,$ $11,$ or $111$ (no longer summand fits, since $1111 \gt 1000$) and the target into $1000.$ If $x,$ $y,$ $z$ count the summands of each size, then $x + 11y + 111z = 1000$ and $n = x + 2y + 3z.$ Subtracting gives $$n = 1000 - 9(y + 12z),$$ so the possible $n$ correspond exactly to the attainable values of $v = y + 12z.$

The constraints are $z \le 9$ and $0 \le y \le \frac{1000 - 111z}{11}$ (then $x \ge 0$ follows). For $z = 0, 1, \ldots, 9$ the value $v = y + 12z$ ranges over the intervals $[0, 90],$ $[12, 92],$ $[24, 94],$ $[36, 96],$ $[48, 98],$ $[60, 100],$ $[72, 102],$ $[84, 104],$ $[96, 106],$ and $\{108\}$ (when $z = 9,$ only $y = 0$ fits). Their union is every integer from $0$ to $106$ together with $108;$ only $107$ is unattainable.

So $v$ takes $107 + 1 = 108$ values, and $n = 1000 - 9v$ takes $108$ values.