Consider an N by N matrix, $\mathcal{M}$, for which $$ \mathcal{M}\phi^i_1=\lambda_i\phi_1 $$ and $$ (\phi^i_2)^T\mathcal{M}=\lambda_i\phi_2 $$
where $\lambda_i$($i=1,..,N$) are real eigenvalues of the matrix, $\phi_1^i$ and $\phi_2^i$ are column matrices for which we get the corresponding eigenvalues $\lambda_i$.
I want to normalize $\phi_1^i$ and $\phi_2^i$ so that $$ (\phi_1^i)^T\phi^j_2 = \delta^{ij} $$
I'm not quite sure how to show this. I understand I have to devide out the eigenvalues somewhere.
This is not a complete answer. We will first assume that the matrix is not deficient and note that $(\phi_1^T \phi_2)_{i j} = (\phi_1^i)^T \cdot \phi_2^j$. This will clearly not be possible if any of the eigenvalues is zero. This is because the row space of $\mathcal{M}$ (denote this by $R$), is perpendicular to the null space of $\mathcal{M}$. $R$ is also the column space of $\mathcal{M}^T$, and we know that the left eigenvectors are in the column space of $\mathcal{M}^T$. So any zero eigenvector will be perpendicular to every column vector of the matrix $\phi_2^i$, and thus $(\phi_1^i)^T \phi_2^j$ will have a row of zeroes.
What you are asking for seems like it may not be possible in the general case. Your problem reduces to showing that there exists matrices $\phi_1$ and $\phi_2$, consisting of left and right eigenvectors of some matrix $\mathcal{M}$, such that $\phi_1^T = \phi_2^{-1}$. Luckily, what you want seems possible in many cases because the left and right eigenvectors of $\mathcal{M}$ of different eigenvalues are necessarily orthogonal. That is, if $\vec{u}^T\mathcal{M} = \lambda \vec{u}^T$ and $\mathcal{M}\vec{v} = \mu \vec{v}$ and $\lambda \neq \mu$, then $\vec{v} \cdot \vec{u} = 0$. Suppose the eigenvalues of $\mathcal{M}$ are nonzero and are $\lambda_1, \cdots \lambda_N$. If the corresponding left-eigenvectors are $\vec{u}_1, \vec{u}_2, \cdots \vec{u}_N$ and the right-eigenvectors are $\vec{v}_1, \vec{v}_2, \cdots, \vec{v}_N$ (in the order of the eigenvalues), then choosing $$\phi_1^i = \frac{1}{(\vec{v}_i \cdot \vec{u}_i)^{1/2}}\vec{v}_i, \text{ }\phi_2^i = \frac{1}{(\vec{v}_i \cdot \vec{u}_i)^{1/2}}\vec{u}_i$$ will achieve the normalization you desire if the eigenvalues are all distinct.
To prove that this works, we will make use of a lemma: If $\vec{u}$ and $\vec{v}$ are left and right eigenvectors of $\mathcal{M}$ which has distinct eigenvalues, they are orthogonal if and only if they do not correspond to the same eigenvalue.
$(\Leftarrow)$ Let $\vec{u}$ correspond to eigenvector $\lambda$, and $\vec{v}$ to correspond to eigenvector $\mu$. Consider the quantity $x = \vec{u}^T \mathcal{M} \vec{v}$. We must have $$x = \vec{u}^T \mathcal{M} \vec{v} = \lambda (\vec{u} \cdot \vec{v}) = \mu (\vec{u} \cdot \vec{v})$$ and since $\lambda \neq \mu$ by assumption, we must have $\vec{u} \cdot \vec{v} = 0$.
$(\Rightarrow)$ Let $\vec{u}$ and $\vec{v}$ correspond to the same eigenvalue $\lambda$. Suppose that $\vec{u} \perp \vec{v}$, i.e. $\vec{u} \cdot \vec{v} = 0$. From the reverse direction, we know $\vec{u}$ is perpendicular to all the other eigenvectors as no eigenvectors of $\mathcal{M}$ share the same eigenvalue. But the union of all eigenspaces for any non-deficient matrix is $\mathbb{R}^n$, and this in particular means that $\vec{u} \in (\vec{v} \cup \vec{v}^\perp)^\perp = (\mathbb{R}^n)^\perp \implies \vec{u} = \vec{0}$, which cannot be possible.
This implies that the diagonal elements of $\phi_1^T \phi_2$ will be nonzero, and are $\phi_1^i \cdot \phi_2^i = (\vec{v}_i \cdot \vec{u}_i) / (\vec{v}_i \cdot \vec{u}_i) = 1$, and the nondiagonal elements will be zero. It also eliminates the possibility where $\vec{v}_i \cdot \vec{u}_i = 0$, as this cannot be possible by our lemma.
Be careful, however: all eigenvalues being distinct is a sufficient condition. I can think of one obvious case where this is possible when not all eigenvalues are distinct, which occurs when $\mathcal{M}$ is symmetric (and has eigenvalues of multiplicity greater than 1). Then the normalization you want will always be possible so long as $\mathcal{M}$ is invertible.