All The Numbers Are Connected!

Let's start with this: $\begin{array}{|ccl|} \hline \text{The number of 1's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 2's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 3's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 4's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 5's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 6's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 7's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 8's in this box is} & \text{\_\_\_\_\_} \\\text{The number of 9's in this box is} & \text{\_\_\_\_\_} \\ \hline \end{array}$

So the first thing to do is to just do the question without considering the actual answers themselves:

$\begin{array}{|ccl|} \hline \text{The number of 1's in this box is} & \text{1} \\\text{The number of 2's in this box is} & \text{1} \\\text{The number of 3's in this box is} & \text{1} \\\text{The number of 4's in this box is} & \text{1} \\\text{The number of 5's in this box is} & \text{1} \\\text{The number of 6's in this box is} & \text{1} \\\text{The number of 7's in this box is} & \text{1} \\\text{The number of 8's in this box is} & \text{1} \\\text{The number of 9's in this box is} & \text{1} \\ \hline \end{array}$

Ok, but we now have a total of 10 1's in the box. We could put in 10 as the number of 1's, but the question specifies that only single digits are allowed. So we'll do this for now:

$\begin{array}{|ccl|} \hline \text{The number of 1's in this box is} & \text{9} \\\text{The number of 2's in this box is} & \text{1} \\\text{The number of 3's in this box is} & \text{1} \\\text{The number of 4's in this box is} & \text{1} \\\text{The number of 5's in this box is} & \text{1} \\\text{The number of 6's in this box is} & \text{1} \\\text{The number of 7's in this box is} & \text{1} \\\text{The number of 8's in this box is} & \text{1} \\\text{The number of 9's in this box is} & \text{2} \\ \hline \end{array}$

Now, you can see that there are 2 twos in the box and if we put a 2 in the twos section we'll get three twos and therefore two threes, like so:

$\begin{array}{|ccl|} \hline \text{The number of 1's in this box is} & \text{9} \\\text{The number of 2's in this box is} & \text{3} \\\text{The number of 3's in this box is} & \text{2} \\\text{The number of 4's in this box is} & \text{1} \\\text{The number of 5's in this box is} & \text{1} \\\text{The number of 6's in this box is} & \text{1} \\\text{The number of 7's in this box is} & \text{1} \\\text{The number of 8's in this box is} & \text{1} \\\text{The number of 9's in this box is} & \text{2} \\ \hline \end{array}$

And now we can see that there are only 6 1's left, so we can just move the 2 that's in the 9's spot:

$\begin{array}{|ccl|} \hline \text{The number of 1's in this box is} & \text{6} \\\text{The number of 2's in this box is} & \text{3} \\\text{The number of 3's in this box is} & \text{2} \\\text{The number of 4's in this box is} & \text{1} \\\text{The number of 5's in this box is} & \text{1} \\\text{The number of 6's in this box is} & \text{2} \\\text{The number of 7's in this box is} & \text{1} \\\text{The number of 8's in this box is} & \text{1} \\\text{The number of 9's in this box is} & \text{1} \\ \hline \end{array}$

Now you can see that everything holds without descending to chaos, as was not the case with quite a few people who tried to solve this problem i.e. my classmates.

And $6 + 3 + 2 + 1 + 1 + 2 + 1 + 1 + 1 = \boxed{18}$

There are 9 statements each containing 2 digits (note each blank is filled by a single digit number). Thus there are 18 digits total, so the sum is $\boxed{18}.$

10+1+1+1+1+1+1+1+1+1 Since there was no restrictions on 0 There are 10 1s including the one in the statement.