I solved it even without any calculations. But who can prove a formula for this task? I'm realy curios.

First you need to prove that digit root of A > 0 is A mod 9 (but not in [0..9), but in [1..9]).

After this we look at number of n/2-digit numbers with square root  equal to 0, 1, .. 9, and sum squares of those numbers, because for every left part the right part may become any value with same digit root.

Obviously, there's only 1 n/2 digit number with 0 square root - that's 000..00 (n/2 times).

As for any other Q square root, we calculate the number of items in the following sequence: Q, Q+9, Q+18, .. Qi, such that Qi < 10^n and i is maximal.

Easy to see, this is (10^(n/2) - 1 - Q) div 9 + 1, and this value doesn't depend on Q (for 1 <= Q <= 9) and is equal to (10^(n/2) - 1) div 9 = 111..11 (n/2 times); let's denote this value with W and don't forget that we take squares, not numbers themselves;

This leads to the formula: F(0)+F(1)+..+F(9) = F(0) + 9 * W * W = 1 + 9 * sqr((10^(n/2) - 1) div 9)