How many pairs (M, N) of sets of size n have M + N = {0, 1, ..., n^2-1}?

Manfred Schroeder, in *Number Theory in Science and Communication*, 4th edition, asks (p. 27): find all pairs of sets $(M,N)$, each of size 10, such that
$$ M + N = \{ m + n : m \in M, n \in N \} $$
is the set $\{ 0, \ldots, 99 \}$. One example is, of course, the pair $M = \{ 0, 1, \ldots 9 \}$ and $N = \{ 0, 10, \ldots, 90 \}$. Of course there's nothing special about 10 here, and one can instead look for pairs $(M,N)$ such that $|M| = |N| = n$ and $M + N = \{0, 1, \ldots, n^2-1\}$.

This reminded me of the following problem: are there two six-sided dice, other than the standard ones, such that the probabilities of obtaining $2, 3, \ldots, 12$ are the same as on the standard dice? The solution I know is as follows: this is equivalent to finding polynomials $f(z), g(z)$ with positive coefficients, with $f(1) = g(1) = 6$, such that $$ f(z) g(z) = z^2 + 2z^3 + 3z^4 + 4z^5 + 5z^6 + 6z^7 + 5z^8 + 4z^9 + 3z^{10} + 2z^{11} + z^{12}$$ Then $f$ and $g$ are the generating functions (polynomials) of the two dice, and $f \cdot g$ the generating function of the possible sums, counted with multiplicity.

Now, this polynomial of degree $12$ factors as $$ z^2 (z+1)^2 (z^2+z+1)^2 (z^2-z+1)^2 $$ and the only way we can group these factors together to get $f(z)$ and $g(z)$ as desired is to take $$f(z) = z(z+1)(z^2+z+1) = 1+2z^2+2z^3+z^4, g(z) = z(z+1)(z^2+z+1)(z^2-z+1)^2 = z+z^3+z^4+z^5+z^6+z^8$$ That is, two dice labelled $(1,2,2,3,3,4)$ and $(1,3,4,5,6,8)$ have the same probabilities of each outcome as the standard dice. (See Sicherman dice in Wikipedia.)

So I tried to do something analogous for Schroeder's problem: factor $1+z+z^2+\cdots+z^{99} = (z^{100}-1)/(z-1)$ into two factors $f(z), g(z)$ with $f(1) = g(1) = 10$ and all coefficients positive. (Factoring $z^{100}-1$ isn't too hard, with the help of the theory of cyclotomic polynomials.) But this method of solution doesn't seem to work. How do we put the factors back together? For example we can take
$$ f(z) = 1 + z + z^2 + z^3 + z^4 + z^{50} + z^{51} + z^{52} + z^{53} + z^{54}, g(z) = 1 + z^5 + z^{10} + z^{15} + z^{20} + z^{25} + z^{30} + z^{35} + z^{40} + z^{45} $$
which I found by experiment. But I can't see how to find all the ``good'' factorizations, since a lot of the irreducible factors of $1+z+z^2 + \cdots + z^{99}$ have negative signs. Presumably someone a bit more comfortable with cyclotomic polynomials can answer Schroeder's question, or the (probably not much simpler) question of finding the *number* of pairs of sets $(M,N)$ having this splitting property.