Justine has two fair dice, one with sides labeled $1,2,\ldots, m$ and one with sides labeled $1,2,\ldots, n.$ She rolls both dice once. If $\tfrac{3}{20}$ is the probability that at least one of the numbers showing is at most 3, find the sum of all distinct possible values of $m+n$.

[i]Proposed by Justin Stevens[/i]
1. We start by using complementary counting to find the probability that neither of the numbers showing is at most 3. This means both dice show numbers greater than 3.
2. The total number of outcomes when rolling two dice is \( mn \).
3. The number of outcomes where both dice show numbers greater than 3 is \( (m-3)(n-3) \).
4. Given that the probability of at least one of the numbers showing is at most 3 is \( \frac{3}{20} \), the probability that neither number is at most 3 is \( 1 - \frac{3}{20} = \frac{17}{20} \).
5. Therefore, we have the equation:
   \[
   \frac{(m-3)(n-3)}{mn} = \frac{17}{20}
   \]
6. Cross-multiplying to clear the fraction, we get:
   \[
   20(m-3)(n-3) = 17mn
   \]
7. Expanding and simplifying, we have:
   \[
   20mn - 60m - 60n + 180 = 17mn
   \]
8. Rearranging terms, we get:
   \[
   3mn - 60m - 60n + 180 = 0
   \]
9. Dividing the entire equation by 3, we obtain:
   \[
   mn - 20m - 20n + 60 = 0
   \]
10. To factor this equation, we use Simon's Favorite Factoring Trick by adding and subtracting 400:
    \[
    mn - 20m - 20n + 60 + 400 - 400 = 0
    \]
    \[
    (m-20)(n-20) = 340
    \]
11. We now find the factor pairs of 340:
    \[
    340 = 1 \times 340, \quad 2 \times 170, \quad 4 \times 85, \quad 5 \times 68, \quad 10 \times 34, \quad 17 \times 20
    \]
12. Each factor pair \((a, b)\) corresponds to \((m-20, n-20)\) or \((n-20, m-20)\). Therefore, we have:
    \[
    (m, n) = (21, 360), (22, 190), (24, 105), (25, 88), (30, 54), (37, 40)
    \]
13. Summing all distinct possible values of \(m+n\):
    \[
    21 + 360 = 381
    \]
    \[
    22 + 190 = 212
    \]
    \[
    24 + 105 = 129
    \]
    \[
    25 + 88 = 113
    \]
    \[
    30 + 54 = 84
    \]
    \[
    37 + 40 = 77
    \]
    \[
    381 + 212 + 129 + 113 + 84 + 77 = 996
    \]

The final answer is \(\boxed{996}\).