Counting danger

First, we find the count $n_a$ and the sum $S_a$ of all four-digit positive integers: $n_a = 9\times 10^3 = 9000$ ; of these 9000 numbers, the average value of the first digit is $5$ , and that of the next three digits it is $4\tfrac12$ , so that the average value of these numbers is $5(4\tfrac12)(4\tfrac12)(4\tfrac12) = 5499\tfrac12,$ so that the sum is $S_a = 9000 \times 5499\tfrac12 = 49\,495\,500.$

We will subtract out all number that have only odd digits. They are of the form $abcd$ where $a,b,c,d \in \{1,3,5,7,9\}$ . $n_o = 5^4 = 625;$ their average value is $5555$ , so $S_o = 625 \times 5555 = 3\,471\,875.$

Likewise, the numbers with only even digits. They are of the form $abcd$ with $a \in \{2,4,6,8\}$ and $b,c,d \in \{0,2,4,6,8\}$ . Thus $n_e = 4\cdot 5^3 = 500,$ and since their average value is $5444$ , we have $S_e = 500 \times 5444 = 2\,722\,000.$

The answers are therefore $n = n_a - n_o - n_e = 9000 - 625 - 500 = 7875,$ $S = S_a - S_o - S_e = 49\,495\,500 - 3\,471\,875 - 2\,722\,000 = 43\,301\,625,$ $S + n = \boxed{43\,309\,500}.$