Beyond Brauer's theorem Brauer's classic theorem states that any character of a finite group can be expressed as a linear combination with integer coefficients of characters induced by linear characters of p-elementary subgroups.
This is an improvement over Artin's theorem, in which the linear combination is with rational coefficients of characters induced by characters of cyclic subgroups.
So we get integer coefficients at the cost of using more complicated subgroups.
It is reasonable to try to improve Brauer's theorem, taking the coeffients to be positive integers. But this doesn't work: there are characters that can't be expressed as linear combination with positive coefficients of characters induced from elementary subgroups, even if we let the coefficients be real numbers (this is exercise 10.5 in Serre's book on representation theory).
In fact, Brauer's result is the best possible in a very concrete sense:


*

*The only groups that have the property that all of its characters are expressible with non-negative coefficients in Brauer's theorem are solvable groups.

*p-elementary subgroups is the minimal family of subgroups for which Brauer's theorem holds (Green).

What I was wondering is, if we replace p-elementary subgroups by a
  bigger (and more complicated) family of subgroups, much like Brauer
  had to let go of cyclic groups, can we get a similar theorem with positive
  coefficients that works for all finite groups?

$\require{AMScd}$
\begin{CD}
    \mathrm{cyclic} @>>> \mathrm{elementary} @>>> \mathrm{?} \\
    @V  V V\ @VV V\ @VV V\\
    \mathbb{Q} @. \mathbb{Z} @. \mathbb{N}
    \end{CD}
 A: A non-zero non-negative linear combination of characters induced from proper subgroups will have degree greater than $1$, so if you want to express every character of a finite group $G$ as a non-negative linear combination of characters induced from subgroups, then you must use $G$ itself as one of those subgroups, or you could never get the trivial character of $G$, or any other linear character.
Taketa's theorem doesn't really have anything to do with the "elementary subgroup" aspect of Brauer's theorem. It considers characters induced from linear characters of arbitrary subgroups.
A: Jeremy Rickard's answer is perfectly correct, of course. It is also easy to see (by an argument close to Taketa's) that if $G$ is a finite simple group, and $\chi$ is a faithful (not assumed  irreducible, though in fact it will be irreducible) complex character of minimal degree of $G$, then $\chi$ is not a non-negative integer combination of characters induced from proper subgroups of $G$. For if $\chi = \sum_{\mu} {\rm Ind}_{H(\mu)}^{G}(\mu),$ where $\mu$ is a character of the proper subgroup $H(\mu)$ of $G$, then there must be a character $\mu$ such that ${\rm Ind}_{H(\mu)}^{G}(\mu)$ is not a multiple of the trivial character. Then (using Frobenius reciprocity)
$\theta = {\rm Ind}_{H(\mu)}^{G}(1) -1$ is a character whose kernel is a normal subgroup of $G$ contained in $H(\mu)$, so $\theta$ is faithful. But certainly $\theta$ has smaller degree than $\chi$, contrary to the minimality of $\chi(1)$.
While they do not address non-negativity issues, there are interesting books on so-called Explicit Brauer Induction ( separate books by R. Boltje and V. Snaith taking a "dual" viewpoints) which discuss minimal families which can be used when seeking induction theorems which commute with restriction to subgroups (in Boltje's case) or induction from subgroups (in Snaith's case).
In a related direction, I remember being present at a discussion where Walter Feit stated that it is the case that whenever $\chi$ is an irreducible complex character of a finite group $G$, there is a finite overgroup $H$ of $G$ and a monomial irreducible character $\theta$ of $H$ such that $\chi$ occurs as a constituent of ${\rm Res}^{H}_{G}(\theta)$. I have no idea how such a statement would be proved, and it may require the classification of finite simple groups. It is related to generalizing in some way to arbitrary finite groups a Theorem of E.C. Dade that every finite solvable group $G$ can be embedded in an $M$-group ( an $M$-group is a finite group all of whose irreducible characters are induced from linear characters of (not necessarily proper) subgroups).
A: Here is another way to think about this, where we consider nonnegative $real$ coefficients, and we do not restrict ourselves to integer coefficients as in Geoff Robinson's post.
If $\chi \in {\rm Irr}(G)$ satisfies $\chi = \sum_\mu a_\mu \mu^G$, where $\mu$ runs over some collection of characters of proper subgroups of $G$ and the coefficients $a_\mu$ are positive (real) numbers, then there exists a positive integer $m$ such that $m\chi$ is induced from a proper subgroup. (In the language of the paper of Le $et\,al$ (J. of Algebra 374 (2013), $\chi$ is "multiply imprimitive".) To see this, note that if $a_\mu > 0$ then $\mu^G$ can have no irreducible constituent other than $\chi$ because there can be no cancellation on account of the assumption that none of the coefficients is negative. It follows that $\mu^G = m\chi$ for some integer $m$, as claimed.
The paper referred to above shows that if every nonlinear irreducible character of a group is multiply imprimitive, then $G$ is solvable. The proof uses the classification of simple groups. 
A stronger property is that $\chi$ is "multiply monomial", which means that some multiple of $\chi$ is induced from a linear character. There is an elementary argument that shows that if every irreducible of $G$ is multiply monomial, then $G$ cannot be perfect. In fact, we argue that in a perfect group $G$, a nonlinear irreducible character of minimum degree cannot be multiply monomial.
Proof: Say $m\chi = \mu^G$, where $\mu \in {\rm Irr}(H)$ is linear, and note that $H < G$. Now write $(1_H)^G = 1_G + \sum a_\psi \psi$, where the $\psi$ are distinct nonprincipal and hence nonlinear irreducible characters and the coefficients $a_\psi$ are positive integers. Also, the set of $\psi$ in this sum is not empty. Then
$$
m\chi(1) = |G:H| = 1 + \sum a_\psi \psi(1) \ge 1 + \chi(1) \sum a_\psi\,,
$$
where the inequality holds because $\psi(1) \ge \chi(1)$ for all $\psi$. Then $m > \sum a_i > 0$, and we have $m^2 > (\sum a_i)^2$, and so
$m^2 \ge 1 + (\sum a_i)^2$. Now 
$$
m^2 = [\mu^G,\mu^G] \le [1_H^G,1_H^G] =
1 + \sum a_i^2 \le 1 + (\sum a_i)^2 \le m^2\,,
$$
Equality thus holds throughout, and thus $m^2$ exceeds the positive square integer $(\sum a_i)^2$ by $1$. This is a contradiction.
