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A triple of vertices in a graph is a \emph{frustrated triangle} if it induces an odd number of edges. We study the set $F_n\subset[0,\binom{n}{3}]$ of possible number of frustrated triangles $f(G)$ in a graph $G$ on $n$ vertices. We prove that about two thirds of the numbers in $[0,n^{3/2}]$ cannot appear in $F_n$, and we characterise the graphs $G$ with $f(G)\in[0,n^{3/2}]$. More precisely, our main result is that, for each $n\geq 3$, $F_n$ contains two interlacing sequences $0=a_0\leq b_0\leq a_1\leq b_1\leq \dots \leq a_m\leq b_m\sim n^{3/2}$ such that $F_n\cap(b_t,a_{t+1})=\emptyset$ for all $t$, where the gaps are $|b_t-a_{t+1}|=(n-2)-t(t+1)$ and $|a_t-b_t|=t(t-1)$. Moreover, $f(G)\in[a_t,b_t]$ if and only if $G$ can be obtained from a complete bipartite graph by flipping exactly $t$ edges/nonedges. On the other hand, we show, for all $n$ sufficiently large, that if $m\in[f(n),\binom{n}{3}-f(n)]$, then $m\in F_n$ where $f(n)$ is asymptotically best possible with $f(n)\sim n^{3/2}$ for $n$ even and $f(n)\sim \sqrt{2}n^{3/2}$ for $n$ odd. Furthermore, we determine the graphs with the minimum number of frustrated triangles amongst those with $n$ vertices and $e\leq n^2/4$ edges., Comment: 19 pages, 4 figures, submitted