24 vectors in Leech lattice having scalar product $\frac{1}{4}$ pairwise

Yes, such a configuration exists, even with all inner products positive (as the title of the question requires, even though the text allows either sign).

We shall use the standard normalization of the Leech lattice $\Lambda$ that makes it unimodular, so the $196560$ minimal vectors $(v,v) = 4$, not $32$. Thus we claim that we can choose $24$ vectors $v_i$ from those $196560$ vectors such that $(v_i,v_j) = 1$ if $i \neq j$.

The Gram matrix of those vectors has determinant $3^{23} (3+24\cdot 3) = 3^{26}$. Instead of finding them in some standard model of $\Lambda$, we find a lattice $L$ containing the ${\bf Z}$-span of the $v_i$ with index $3^{13}$, and verify that $L$ is an even unimodular lattice with no vectors of norm $2$; it is known that this suffices to prove $L \cong \Lambda$. I do not know whether the resulting construction of $\Lambda$ is new or equivalent to one of the many previously known ones (though the ingredients are certainly familiar).

We have not yet named the $24$ indices $i$ of our $v_i$. Let us call them $1,2,3,\ldots,23,\infty$, with all but $\infty$ taken mod $23$. We generate $\Lambda$ by the $v_i$ together with $u_\infty := \frac19(v_\infty + \sum_{i \bmod 23} v_i)$ and the $23$ vectors $u_j := \frac13 \sum_{a=0}^{11} v_{a^2+j}$.

It is straightforward to check that all these vectors generate an even lattice $L$. The fact that $L$ is unimodular requires the dimension of ternary quadratic-residue codes, and then checking that $L$ has no vectors of norm $2$ would take rather more work $-$ so instead we write and run a few lines of gp code:

G1 = matrix(24,24,i,j,if(i==j,4,1))
B = matrix(24,24,i,j,if(j==24,1,if(i<24,3*(kronecker(i-j,23)>=0))))
B = concat(9*matid(24), B);
A = matrixqz(B,-1)
A /= 9;
G = A~ * G1 * A
matdet(G)
qfminim(G,,0)

The first line constructs the Gram matrix of the $v_i$; the second line constructs a $24\times 24$ matrix $B$ with columns $9u_i$ and $9u_\infty$; the third line adds the $24$ columns $9v_i$ by extending $B$ by the matrix $9I_{24}$; the next line finds a $24 \times 24$ matrix $A$ whose columns generate the ${\bf Z}$-span of those 48 vectors. We then divide by $9$ and compute the Gram matrix $G$ of this span with respect to the $A$ generators. Finally we verify thet $\det G = 1$ and ask for the number and norm of minimal nonzero vectors (the concluding ",,0" suppresses the output of those minimal vectors). The result is

[196560, 4, [;]]

confirming $L \cong \Lambda$, QED.

P.S. Is it clear that there is no such configuration with more than $24$ vectors? There certainly cannot be 26, because their Gram matrix would be invertible even modulo $2$; but conceivably a 25th vector might be added if we allow both $+1$ and $-1$ inner products.

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Added later: Marek Mitros notes in a comment that the inner products $(v_i,v_j) = 3 \delta_{ij} + 1$ are realized by the $24$ vectors of shape $12^{-1/2} (-5, 1^{23})$ in ${\bf R}^{24}$. In those coordinates my description of $\Lambda$ amounts to the following simpler construction, which should already be known $-$ indeed it seems to be the same as doubling the lattice obtained by "construction B$_3$" from the ternary QR (quadratic-residue) code of length $24$, see pages 148--149 (Chapter 5, $\S\S$5.5--5.7) in Conway and Sloane's SPLAG = Sphere Packings, Lattices and Groups.

Let $e_i$ be the $24$ pairwise orthogonal unit vectors (with $i=1,2,\ldots,23,\infty$ as before), and let $f_i$ be the rows of a $24 \times 24$ Hadamard matrix constructed using quadratic residues mod $23$: $f_{ij} = -1$ unless $i,j \neq \infty$ and $i-j$ is a square mod $23$ (including zero), in which case $f_{ij} = +1$. Then $12^{1/2}\Lambda$ is generated by the $24+24$ vectors $6(e_i + e_\infty)$ and $f_i - 6e_\infty$ (NB $i = \infty$ is allowed), with $v_i = -(f_\infty - 6e_\infty) - 6(e_i+e_\infty)$.

Here's the gp code:

E = 6 * matrix(24,24,i,j,(i==j)+(i==24));
F = matrix(24,24,i,j, if((i<24) && (j<24) && kronecker(i-j,23) >= 0, 1, -1));
for(j=1,24,F[24,j]-=6);
A = matrixqz(concat(E,F),-1)
G = A~ * A / 12
matdet(G) \\ = 1
qfminim(G,,0)

which again outputs

[196560, 4, [;]]

SPLAG reports that all that this construction requires of the QR code is that it be self-dual with parameters $[24,12,9]$, and thus that the ternary Pless (double circulant) code of the same length works as well; this should give an inequivalent configuration of $24$ Leech vectors of norm $4$ with all inner products equal $1$.

Here are 24 vectors in classic Leech lattice having pairwise scalar product 8. I obtained it from $(-5, 1^{23})$ using this.

$\begin{bmatrix} -1& 1& 1& 1& 1&-1& 1&-1& 1& 1&-1&-1& 1& 1&-1&-1& 1&-1& 1&-1&-1&-1&-1&-3 \\ \dots\\ 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1& 1&-3 \end{bmatrix}$

The dots $\dots$ should be replaced by 22 vectors obtained from the first one by permutation $(1,2,...,23)$.

As a consequence we can obtain following useful

Fact

For each element $p$ of order $23$ in $M_{23}$ there exist dodecad $d$ such that dodecads $d, pd, p^2d,...,p^{22}d$ intersect in six points pairwise.