Theorem If $m$ is relatively prime to $30$, then $N=3$ works.
I'll use additive notation. The cyclic group of order $n$ is denoted $C(n)$, the generators of this cyclic group are denoted $U(n)$.
Lemma Let $p \geq 7$ be prime and let $X$, $Y$ and $Z$ be subsets of $C(p)$ of size at least $(p-1)/2$. Then $0 \in X+Y+Z$.
Proof By the Cauchy-Davenport theorem (thanks quid!), we have $|X+Y+Z| \geq \min(3 (p-1)/2 -2,p)$, which equals $p$ once $p \geq 7$. $\square$
We now prove the Theorem by induction on $m$; the case $m=1$ is vacuously true. Let $p$ be a prime dividing $m$ and let $\pi$ be the projection map $C(m) \to C(m/p)$. Let $A \subset U(m)$ so that $U(m)$ is the disjoint union of $A$ and $-A$. We want to show that $0 \in A+A+A$.
For $b \in U(m/p)$, we have $|\pi^{-1}(b) \cap A|+|\pi^{-1}(-b) \cap A| = p-1$ or $p$; the first case occurs when $GCD(p,m/p)=1$ and the second case occurs otherwise. Choose a subset $B$ of $U(m/p)$ so that $U(m/p)$ is the disjoint union of $B$ and $-B$ and, for each $b \in B$, we have $|\pi^{-1}(b) \cap A| \geq (p-1)/2$.
Inductively, we can find $b_1$, $b_2$ and $b_3$ in $B$ (not necessarily distinct) so that $b_1+b_2+b_3=0$. Choose lifts $c_i$ of the $b_i$ to $C(m)$ so that $c_1+c_2+c_3=0$. Let $X_i$ be $(\pi^{-1}(b_i) \cap A) - c_i$ for $i=1$, $2$, $3$. Use the Lemma to find $x_1$, $x_2$, $x_3$ with $x_i \in X_i$ and $x_1+x_2+x_3=0$. Then $c_i+x_i \in \pi^{-1}(b_i) \cap A \subseteq A$ and $\sum (c_i+x_i) = 0$. $\square$
As noted in comments below, $12$ always works. Indeed, if we want to have a single $N$ for all $\Phi$, then $12$ is the best possible because $m=12$, $\Phi = \{1,7 \}$ forces $6|N$ and $m=12$, $\Phi = \{ 1,5 \}$ forces $4 | N$. However, it seems to me that a better statement (which I will now prove) is that, for any $\Phi$, either $N=4$ or $N=6$ works.
Once our inductive claim is that either $4$ or $6$ works, the prime $5$ behaves like any other prime, so we may reduce to numbers of the form $2^a 3^b$. Also, if $9 | m$ and the claim holds for $m/3$, then it holds for $m$, as we get to replace the bound $(p-1)/2$ by $p/2$. So we are reduced to $m$ of the form $2^a$ or $2^a \times 3$.
gcousin has already done $2^a$. For $m=2^a \times 3$, consider the map $\pi: C(2^a \times 3) \to C(2^a)$. If all of the fibers of this map have size $0$ or $2$, then we may reduce to the case of $2^a$ as above, since then we have subsets of $C(3)$ of size $2$ in the lemma.
Suppose, then, that there is some $x \in U(2^a)$ such $\pi^{-1}(x) \cap A$ and $\pi^{-1}(-x) \cap A$ nonzero; let $a$ and $a'$ occupy these sets. Then $a+a'$ is $3$-torsion, so $N=6$ works.
Best Answer
In this paper they talk about this problem for 5 instead of 10 roots.
http://www.jstor.org/stable/2323469
EDIT: In view of Todd Trimble's comment, here's a summary of what's in the paper.
Let $f(k,N)$ be the least absolute value of a nonzero sum of $k$ (not necessarily distinct) $N$-th roots of unity. Then
$f(2,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi$ for even $N$, $\pi$ for odd $N$,
$f(3,N)$ is asymptotic to $cN^{-1}$, where $c$ is $2\pi\sqrt3$ for $N$ divisible by 3, $2\pi\sqrt3/3$ otherwise,
$f(4,N)$ is asymptotic to $cN^{-2}$, where $c$ is $4\pi^2$ for even $N$, $\pi^2$ for $N$ odd,
$f(k,N)>k^{-N}$ for all $k,N$,
$f(2s,N)<c_sN^{-s}$ for $N$ even and $s\le10$,
$f(k,N)<c_kN^{-[\sqrt{k-6}]-1}$ for $N$ even and $k>5$, and
If $N$ is twice a prime, and $k<N/2$, then there exists $k'<2k$ such that $f(k',N)\le2k2^{k/2}\sqrt{k!}N^{-k/2}$.
The only result in the paper for 5 roots of unity is (the trivial) $f(5,N)>5^{-N}$, but it is suggested that maybe $f(5,N)>cN^{-d}$ for some $d$, $2\le d\le3$, and some $c>0$.