Let's fill in the dots with positive integers in the decimal system so that the statement in the frame is true:

In this frame, there are exactly ... instances of the digit 0, ... instances of the digit 1, ... instances of the digit 2, ... instances of the digit 3, ... instances of the digit 4, ... instances of the digit 5, ... instances of the digit 6, ... instances of the digit 7, ... instances of the digit 8, and finally ... instances of the digit 9.
Consider the task as solved, and let $A$ denote the number of digits of the numbers written in the dotted places, and $S$ the sum of these numbers. $S$ is nothing more than the total number of digits occurring in the frame, $A+10$. We wrote numbers in ten places in the frame, so $A \geqq 10$. If the numbers written are single-digit - which was taken for granted in most of the submitted solutions - then $A=10$ and thus $S=20$. However, it is possible that we did not write only single-digit numbers. In this case, $A>10$, and at least $(A-10)$ digits are in the tens place. The sum of the numbers written, $A+10$, is then at least $10(A-10)$, from which $A \leqq \frac{110}{9}A$.

The sum of the numbers written, $S$, is therefore either 20 or 21, in the latter case, we wrote exactly one two-digit number in the frame. Let the number of occurrences of the $i$ digit in the frame be $a_{i}$, and first examine the case $S=a_{0}+a_{1}+\ldots+a_{9}=20$.

Since $a_{1}=1$ cannot be, 12. The sum of the other 7 numbers written is thus at most 6, which is not possible.

In the case of $a_{1}=11$, $a_{0}=1$, so out of the 11 one's, $a_{0}$, $a_{1}$, and the pre-written 1 make up four, so 1 goes on 7 of the remaining 8 places. The sum of these 9 numbers is $11+8=19$, so the only digit different from 1 must be 2. Therefore, in the frame, there can be 2 twos, 11 ones, and every other digit can only appear once. The second solution to the problem is thus the following:

In this frame, there is exactly $\mathbf{1}$ zero, $\mathbf{11}$ ones,

2 twos, 1 three, 1 four, 1 five, 1 six,

$\mathbf{1}$ seven, $\mathbf{1}$ eight, and exactly $\mathbf{1}$ nine.

Remarks. 1. The second solution can be written in any base $g>2$ number system, ${ }^{g} 11$ ones mean exactly $(g+1)$ one's digits. (The digits, of course, range from 0 to $(g-1)$.) If $g=2$, then "11 zeros and 100 ones" is the solution.

2. The statement in the frame makes a statement about itself. Such statements - e.g., "I am lying" - have long been used in constructing various logical paradoxes. More about this can be read in László Csirmaz's article "Logical, aren't I?" (KÖMAL, 1982/3, pp. $170-172$).