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Here is an answer (see the last point). It differs from what I had been claiming in my first post. There I was saying that any positive bilinear module $\Lambda$ over $\mathbf Z[\frac 12]$ was representable by some euclidean module $\mathrm{I}_n\otimes\mathbf Z[\frac 12]$. This is true (with $n\leq \text{rk}(\Lambda)+3$), but is of no use in our situation.

Nevertheless the same technics show that any positive bilinear $m$-dimensional module over $\mathbf Z$ is representable by somebody in the same genus than the euclidean module of rank $m+4$. This is proved here in the third point, since it might be of interest.

In terms of matrices, as in the OP, it says that for any symmetric positive definite matrix $M\in\mathrm{Sym}_m(\mathbf Z)$ there exists a symmetric positive definite matrix $G\in\mathrm{Sym}_{m+4}(\mathbf Z)$ with determinant $1$ and at least one odd diagonal entry, and a matrix $Q\in \mathrm{Mat}_{m+4,m}(\mathbf Z)$ such that the following holds : $$Q^t.G.Q=M\ \ \ .$$

Let $M$ be a positive definite symmetric $m\times m$ matrix. Let $\mathrm{M}$ be the bilinear module $(\mathbf{Z}^m,M)$. The question can be rephrased :

What is the smallest value $n$ such that the standard euclidean bilinear module $\mathrm{I}_n$ represents $\mathrm{M}$.

$\blacktriangleright$ First note that in general, such an $n$ doesn't exist. Indeed, a positive definite integral lattice decomposes as the orthogonal sum of indecomposable (for the orthogonal direct sum) lattices in a unique manner. In particular, the only unimodular (i.e. $\det(M)=1$) lattice $\mathrm{M}$ represented by $\mathrm{I}_n$ are isomorphic to $\mathrm{I}_m$.

$\blacktriangleright$ Over $\mathbf Q$, the space $\mathrm{M}\otimes\mathbf Q$ is represented by $\mathrm{I}_{n}\otimes\mathbf Q$ for some $n\leq m+3$. This follows from the Hasse principle (that implies that a rank $4$ positive definite space represents $1$), Witt cancellation, and the fact that, for example, $(\mathrm{M}\otimes\mathbf Q)^{\perp 4}$ is euclidean, as a computation of Hasse-Minkowski symbols shows.

$\blacktriangleright$ It follows that over $\mathbf Z$, any bilinear module $\mathrm{M}$ is represented by some bilinear module $\mathrm{N}$ lying in the genus of $\mathrm{I}_{n}$, for some $n\leq m+4$ (the addition of a one dimensional module $\mathrm{I}_1$ might be necessary when $m+3$ is a multiple of $8$, in order to ensure that $\mathrm{N}$ is odd).

$\blacktriangleright$ Finally, here is an answer to the question. Let $M$ be an $m$-dimensional submodule of $\mathrm{I}_n$. Let $P$ be its orthogonal. Let $\pi : \mathrm{I}_n\otimes Q\to M\otimes Q$ be the orthogonal projection. Then $\pi(\mathrm{I}_n)$ is contained in $M^\sharp$ (the dual lattice of $M$ in $M\otimes \mathbf Q$), and there exists a smallest $d$ such that $M^\sharp\subset d^{-1}M$. Let $S$ be the sphere of unitary vectors in $\mathrm{I}_n$. Then the vectors of $d.\pi(S)\subset M$ have squared length smaller than $d^2$. If the number of such vectors is smaller that $\mathrm{card}(S)=2^n$, this means that a vector of squared length $2$ in $I_n$ lies in $P$, so $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-2}\perp <2>$. If once again the same phenomenon occurs, then $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-4}\perp <2,2>$ which is represented by $\mathrm{I}_{n-2}$. So what we get is :

Let $d$ be the smallest integer such that $\mathrm{M}^\sharp\subset \frac 1d\mathrm{M}$. Let $B_d(\mathrm{M})$ denote the number of non-trivial vectors of $M$ in the ball of radius $d$. Let $n_0$ be the floor of $2+\log_2(B_d(\mathrm{M}))$. If $\mathrm{M}$ is represented by $\mathrm{I}_n$ for some $n$, then $\mathrm{M}$ is represented by $\mathrm{I}_{n_0}$.

I hope there are better bounds, but I cannot see how to get them.

Here is an answer (see the last point). It differs from what I had been claiming in my first post. There I was saying that any positive bilinear module $\Lambda$ over $\mathbf Z[\frac 12]$ was representable by some euclidean module $\mathrm{I}_n\otimes\mathbf Z[\frac 12]$. This is true (with $n\leq \text{rk}(\Lambda)+3$), but is of no use in our situation.

Nevertheless the same technics show that any positive bilinear $m$-dimensional module over $\mathbf Z$ is representable by somebody in the same genus than the euclidean module of rank $m+4$. This is proved here in the third point, since it might be of interest.

In terms of matrices, as in the OP, it says that for any symmetric positive definite matrix $M\in\mathrm{Sym}_m(\mathbf Z)$ there exists a symmetric positive definite matrix $G\in\mathrm{Sym}_{m+4}(\mathbf Z)$ with determinant $1$ and at least one odd diagonal entry, and a matrix $Q\in \mathrm{Mat}_{m+4,m}(\mathbf Z)$ such that the following holds : $$Q^t.G.Q=M\ \ \ .$$

Let $M$ be a positive definite symmetric $m\times m$ matrix. Let $\mathrm{M}$ be the bilinear module $(\mathbf{Z}^m,M)$. The question can be rephrased :

What is the smallest value $n$ such that the standard euclidean bilinear module $\mathrm{I}_n$ represents $\mathrm{M}$.

$\blacktriangleright$ First note that in general, such an $n$ doesn't exist. Indeed, a positive definite integral lattice decomposes as the orthogonal sum of indecomposable (for the orthogonal direct sum) lattices in a unique manner. In particular, the only unimodular (i.e. $\det(M)=1$) lattice $\mathrm{M}$ represented by $\mathrm{I}_n$ are isomorphic to $\mathrm{I}_m$.

$\blacktriangleright$ Over $\mathbf Q$, the space $\mathrm{M}\otimes\mathbf Q$ is represented by $\mathrm{I}_{n}\otimes\mathbf Q$ for some $n\leq m+3$. This follows from the Hasse principle (that implies that a rank $4$ positive definite space represents $1$), Witt cancellation, and the fact that, for example, $(\mathrm{M}\otimes\mathbf Q)^{\perp 4}$ is euclidean, as a computation of Hasse-Minkowski symbols shows.

$\blacktriangleright$ It follows that over $\mathbf Z$, any bilinear module $\mathrm{M}$ is represented by some bilinear module $\mathrm{N}$ lying in the genus of $\mathrm{I}_{n}$, for some $n\leq m+4$ (the addition of a one dimensional module $\mathrm{I}_1$ might be necessary when $m+3$ is a multiple of $8$, in order to ensure that $\mathrm{N}$ is odd).

$\blacktriangleright$ Finally, here is an answer to the question. Let $M$ be an $m$-dimensional submodule of $\mathrm{I}_n$. Let $P$ be its orthogonal. Let $\pi : \mathrm{I}_n\otimes Q\to M\otimes Q$ be the orthogonal projection. Then $\pi(\mathrm{I}_n)$ is contained in $M^\sharp$ (the dual lattice of $M$ in $M\otimes \mathbf Q$), and there exists a smallest $d$ such that $M^\sharp\subset d^{-1}M$. Let $S$ be the sphere of unitary vectors in $\mathrm{I}_n$. Then the vectors of $d.\pi(S)\subset M$ have squared length smaller than $d^2$. If the number of such vectors is smaller than $\mathrm{card}(S)=2^n$, this means that a vector of squared length $2$ in $I_n$ lies in $P$, so $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-2}\perp <2>$. If once again the same phenomenon occurs, then $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-4}\perp <2,2>$ which is represented by $\mathrm{I}_{n-2}$. So what we get is :

Let $d$ be the smallest integer such that $\mathrm{M}^\sharp\subset \frac 1d\mathrm{M}$. Let $B_d(\mathrm{M})$ denote the number of non-trivial vectors of $M$ in the ball of radius $d$. Let $n_0$ be the floor of $2+\log_2(B_d(\mathrm{M}))$. If $\mathrm{M}$ is represented by $\mathrm{I}_n$ for some $n$, then $\mathrm{M}$ is represented by $\mathrm{I}_{n_0}$.

I hope there are better bounds, but I cannot see how to get them.

Here is an answer (the last point). It differs from what I had been claiming in my first post. There I was saying that any positive bilinear module over $\mathbf Z[\frac 12]$ was representable by some euclidean module over this ring. This is true, but is of no use.

Nevertheless the same technics show that any positive bilinear $m$-dimensional module over $\mathbf Z$ is representable by somebody in the same genus than the euclidean module of rank $m+4$. This is proved here in the third point, since it might be of interest.

Let $M$ be a positive definite symmetric $m\times m$ matrix. Let $\mathrm{M}$ be the bilinear module $(\mathrm{Z}^m,M)$. The question can be rephrased :

What is the smallest value $n$ such that the standard euclidean bilinear module $\mathrm{I}_n$ represents $\mathrm{M}$.

$\blacktriangleright$ First note that in general, such an $n$ doesn't exist. Indeed, a positive definite integral lattice decomposes as the orthogonal sum of indecomposable (for the orthogonal direct sum) lattices in a unique manner. In particular, the only unimodular (i.e. $\det(M)=1$) lattice $\mathrm{M}$ represented by $\mathrm{I}_n$ are isomorphic to $\mathrm{I}_m$.

$\blacktriangleright$ Over $\mathbf Q$, the space $\mathrm{M}\otimes\mathbf Q$ is represented by $\mathrm{I}_{n}\otimes\mathbf Q$ for some $n\leq m+3$. This follows from the Hasse principle (that implies that a rank $4$ positive definite space represents $1$), Witt cancellation, and the fact that, for example, $(\mathrm{M}\otimes\mathbf Q)^{\perp 4}$ is euclidean, as a computation of Hasse-Minkowski symbols shows.

$\blacktriangleright$ It follows that over $\mathbf Z$, any bilinear module $\mathrm{M}$ is represented by some bilinear module $\mathrm{N}$ lying in the genus of $\mathrm{I}_{n}$, for some $n\leq m+4$ (the addition of a one dimensional module $\mathrm{I}_1$ might be necessary when $m+3$ is a multiple of $8$, in order to ensure that $\mathrm{N}$ is odd).

$\blacktriangleright$ Finally, here is an answer to the question. Let $M$ be an $m$-dimensional submodule of $\mathrm{I}_n$. Let $P$ be its orthogonal. Let $\pi : \mathrm{I}_n\otimes Q\to M\otimes Q$ be the orthogonal projection. Then $\pi(\mathrm{I}_n)$ is contained in $M^\sharp$ (the dual lattice of $M$ in $M\otimes \mathbf Q$), and there exists a smallest $d$ such that $M^\sharp\subset d^{-1}M$. Let $S$ be the sphere of unitary vectors in $\mathrm{I}_n$. Then the vectors of $d.\pi(S)\subset M$ have squared length smaller than $d^2$. So what we get is :

Let $d$ be the smallest integer such that $\mathrm{M}^\sharp\subset \frac 1d\mathrm{M}$. Let $B_d(\mathrm{M})$ denote the number of non-trivial vectors of $M$ in the ball of radius $d$. Let $n_0$ be the floor of $\log_2(B_d(\mathrm{M}))$. If $\mathrm{M}$ is represented by $\mathrm{I}_n$ for some $n$, then $\mathrm{M}$ is represented by $\mathrm{I}_{n_0}$.

Here is an answer (see the last point). It differs from what I had been claiming in my first post. There I was saying that any positive bilinear module $\Lambda$ over $\mathbf Z[\frac 12]$ was representable by some euclidean module $\mathrm{I}_n\otimes\mathbf Z[\frac 12]$. This is true (with $n\leq \text{rk}(\Lambda)+3$), but is of no use in our situation.

Nevertheless the same technics show that any positive bilinear $m$-dimensional module over $\mathbf Z$ is representable by somebody in the same genus than the euclidean module of rank $m+4$. This is proved here in the third point, since it might be of interest.

In terms of matrices, as in the OP, it says that for any symmetric positive definite matrix $M\in\mathrm{Sym}_m(\mathbf Z)$ there exists a symmetric positive definite matrix $G\in\mathrm{Sym}_{m+4}(\mathbf Z)$ with determinant $1$ and at least one odd diagonal entry, and a matrix $Q\in \mathrm{Mat}_{m+4,m}(\mathbf Z)$ such that the following holds : $$Q^t.G.Q=M\ \ \ .$$

Let $M$ be a positive definite symmetric $m\times m$ matrix. Let $\mathrm{M}$ be the bilinear module $(\mathbf{Z}^m,M)$. The question can be rephrased :

What is the smallest value $n$ such that the standard euclidean bilinear module $\mathrm{I}_n$ represents $\mathrm{M}$.

$\blacktriangleright$ First note that in general, such an $n$ doesn't exist. Indeed, a positive definite integral lattice decomposes as the orthogonal sum of indecomposable (for the orthogonal direct sum) lattices in a unique manner. In particular, the only unimodular (i.e. $\det(M)=1$) lattice $\mathrm{M}$ represented by $\mathrm{I}_n$ are isomorphic to $\mathrm{I}_m$.

$\blacktriangleright$ Over $\mathbf Q$, the space $\mathrm{M}\otimes\mathbf Q$ is represented by $\mathrm{I}_{n}\otimes\mathbf Q$ for some $n\leq m+3$. This follows from the Hasse principle (that implies that a rank $4$ positive definite space represents $1$), Witt cancellation, and the fact that, for example, $(\mathrm{M}\otimes\mathbf Q)^{\perp 4}$ is euclidean, as a computation of Hasse-Minkowski symbols shows.

$\blacktriangleright$ It follows that over $\mathbf Z$, any bilinear module $\mathrm{M}$ is represented by some bilinear module $\mathrm{N}$ lying in the genus of $\mathrm{I}_{n}$, for some $n\leq m+4$ (the addition of a one dimensional module $\mathrm{I}_1$ might be necessary when $m+3$ is a multiple of $8$, in order to ensure that $\mathrm{N}$ is odd).

$\blacktriangleright$ Finally, here is an answer to the question. Let $M$ be an $m$-dimensional submodule of $\mathrm{I}_n$. Let $P$ be its orthogonal. Let $\pi : \mathrm{I}_n\otimes Q\to M\otimes Q$ be the orthogonal projection. Then $\pi(\mathrm{I}_n)$ is contained in $M^\sharp$ (the dual lattice of $M$ in $M\otimes \mathbf Q$), and there exists a smallest $d$ such that $M^\sharp\subset d^{-1}M$. Let $S$ be the sphere of unitary vectors in $\mathrm{I}_n$. Then the vectors of $d.\pi(S)\subset M$ have squared length smaller than $d^2$. If the number of such vectors is smaller that $\mathrm{card}(S)=2^n$, this means that a vector of squared length $2$ in $I_n$ lies in $P$, so $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-2}\perp <2>$. If once again the same phenomenon occurs, then $\mathrm{M}$ is in fact represented by $\mathrm{I}_{n-4}\perp <2,2>$ which is represented by $\mathrm{I}_{n-2}$. So what we get is :

Let $d$ be the smallest integer such that $\mathrm{M}^\sharp\subset \frac 1d\mathrm{M}$. Let $B_d(\mathrm{M})$ denote the number of non-trivial vectors of $M$ in the ball of radius $d$. Let $n_0$ be the floor of $2+\log_2(B_d(\mathrm{M}))$. If $\mathrm{M}$ is represented by $\mathrm{I}_n$ for some $n$, then $\mathrm{M}$ is represented by $\mathrm{I}_{n_0}$.

I hope there are better bounds, but I cannot see how to get them.

I have a candidate : $m\leq 3n+4$.

I don't have enough time to write the proof by now ... If no better bound comes up, I'll write it tomorrow.

Just : first do it over $\mathbf Z[\frac{1}{2}]$, this gives the bound (over this ring, any positive definite rank $n$ module is represented by $\mathrm{I}_{3n+4}$). Then, use the uniqueness of the decomposition of a positive definite bilinear module over $\mathbf Z$ in an orthogonal sum of indecomposable submodules.

Here is an answer (the last point). It differs from what I had been claiming in my first post. There I was saying that any positive bilinear module over $\mathbf Z[\frac 12]$ was representable by some euclidean module over this ring. This is true, but is of no use.

Nevertheless the same technics show that any positive bilinear $m$-dimensional module over $\mathbf Z$ is representable by somebody in the same genus than the euclidean module of rank $m+4$. This is proved here in the third point, since it might be of interest.

Let $M$ be a positive definite symmetric $m\times m$ matrix. Let $\mathrm{M}$ be the bilinear module $(\mathrm{Z}^m,M)$. The question can be rephrased :

What is the smallest value $n$ such that the standard euclidean bilinear module $\mathrm{I}_n$ represents $\mathrm{M}$.

$\blacktriangleright$ First note that in general, such an $n$ doesn't exist. Indeed, a positive definite integral lattice decomposes as the orthogonal sum of indecomposable (for the orthogonal direct sum) lattices in a unique manner. In particular, the only unimodular (i.e. $\det(M)=1$) lattice $\mathrm{M}$ represented by $\mathrm{I}_n$ are isomorphic to $\mathrm{I}_m$.

$\blacktriangleright$ Over $\mathbf Q$, the space $\mathrm{M}\otimes\mathbf Q$ is represented by $\mathrm{I}_{n}\otimes\mathbf Q$ for some $n\leq m+3$. This follows from the Hasse principle (that implies that a rank $4$ positive definite space represents $1$), Witt cancellation, and the fact that, for example, $(\mathrm{M}\otimes\mathbf Q)^{\perp 4}$ is euclidean, as a computation of Hasse-Minkowski symbols shows.

$\blacktriangleright$ It follows that over $\mathbf Z$, any bilinear module $\mathrm{M}$ is represented by some bilinear module $\mathrm{N}$ lying in the genus of $\mathrm{I}_{n}$, for some $n\leq m+4$ (the addition of a one dimensional module $\mathrm{I}_1$ might be necessary when $m+3$ is a multiple of $8$, in order to ensure that $\mathrm{N}$ is odd).

$\blacktriangleright$ Finally, here is an answer to the question. Let $M$ be an $m$-dimensional submodule of $\mathrm{I}_n$. Let $P$ be its orthogonal. Let $\pi : \mathrm{I}_n\otimes Q\to M\otimes Q$ be the orthogonal projection. Then $\pi(\mathrm{I}_n)$ is contained in $M^\sharp$ (the dual lattice of $M$ in $M\otimes \mathbf Q$), and there exists a smallest $d$ such that $M^\sharp\subset d^{-1}M$. Let $S$ be the sphere of unitary vectors in $\mathrm{I}_n$. Then the vectors of $d.\pi(S)\subset M$ have squared length smaller than $d^2$. So what we get is :

Let $d$ be the smallest integer such that $\mathrm{M}^\sharp\subset \frac 1d\mathrm{M}$. Let $B_d(\mathrm{M})$ denote the number of non-trivial vectors of $M$ in the ball of radius $d$. Let $n_0$ be the floor of $\log_2(B_d(\mathrm{M}))$. If $\mathrm{M}$ is represented by $\mathrm{I}_n$ for some $n$, then $\mathrm{M}$ is represented by $\mathrm{I}_{n_0}$.