Here is a family with cardinality continuum. For a nonnegative integer $n$, let the base $2$ expansion of $n$ be $$n = \sum_{k=0}^{\infty} b_k(n) 2^k.$$ So $b_k(n) \in \{ 0, 1 \}$ and is equal to $1$ for all but finitely many $k$.
Now, let $\lambda = (\lambda_0, \lambda_1, \ldots, )$ be any sequence in $\{ 0, 1 \}^{\infty}$ and define $$\chi_{\lambda}(n) = (-1)^{\sum_k b_k(n) \lambda_k}.$$
I claim that $\chi_{\lambda}$ and $\chi_{\mu}$ are almost orthogonal for all $\lambda \neq \mu$.
Let $k$ be an index for which $\lambda_k \neq \mu_k$. For $n = 2^k m + r$ with $0 \leq r < 2^k$, we have $\chi_{\lambda}(n) = \chi_{\lambda}(2^k m) \chi_{\lambda}(r)$. Therefore, $$\sum_{n=2^k m}^{2^k m+2^k-1} \chi_{\lambda}(n) \chi_{\mu}(n) = \chi_{\lambda}(2^k m) \chi_{\mu}(2^k m) \sum_{r=0}^{2^k-1} \chi_{\lambda}(r) \chi_{\mu}(r) = \chi_{\lambda}(2^k m) \chi_{\mu}(2^k m) \prod_{j=0}^{k-1} (1+(-1)^{\lambda_j \mu_j}) = 0.$$ The last equality is because at least one term in the product is $0$, namely $1+(-1)^{\lambda_k \mu_k}$.
So the sum of $\chi_{\lambda} \chi_{\mu}$ is $0$ over intervals of the form $[2^k m, 2^k m + 2^k-1]$, and thus the sum of $\chi_{\lambda} \chi_{\mu}$ over any $[0, N]$ is at most $2^k-1$. $\square$.