If $n$ vectors $a_1, a_2, \cdots , a_n$ are given in $\mathbb{R}^n$ with all lengths at most $1$, then it is not to hard to see that we can put $+$ and $-$ in place of $*$ in the expression $$a_1 * a_2 * \cdots * a_n$$ so that the result will have length at most $\sqrt{n}$.

To see it's enough to choose $\lambda_1, \cdots, \lambda_n$ independently and uniformly from $\{\pm 1\}$. Then, put $X={|\sum_{i=1}^{n}a_i\lambda_i|}^2$. By a simple calculation, you will find that $E(X)=n$. Henee there exist specific $\lambda_1, \cdots, \lambda_n = \pm 1$ with and with $X\leq n$. Taking square roots gives the theorem.

Now I am curious to know, is it possible to generalize this theorem as follows!

Assume $N>n$ vectors $a_1, a_2, \cdots , a_N$ are given in $\mathbb{R}^n$ with all lengths at most $1$. Is it possible to put $+$ and $-$ in place of $*$ in the expression $$a_1 * a_2 * \cdots * a_N ,$$ so that the result will have length at most $\sqrt{n}$.

If $n$ vectors $a_1, a_2, \cdots , a_n$ are given in $\mathbb{R}^n$ with all lengths at most $1$, then it is not to hard to see that we can put $+$ and $-$ in place of $*$ in the expression $$a_1 * a_2 * \cdots * a_n$$ so that the result will have length at most $\sqrt{n}$.

To see this, it's enough to choose $\lambda_1, \cdots, \lambda_n$ independently and uniformly from $\{\pm 1\}$. Then, put $X={|\sum_{i=1}^{n}a_i\lambda_i|}^2$. By a simple calculation, you will find that $E(X)=n$. Hence there exist specific $\lambda_1, \cdots, \lambda_n = \pm 1$ with and with $X\leq n$. Taking square roots gives the theorem.

Now I am curious to know, is it possible to generalize this theorem as follows?

Assume $N>n$ vectors $a_1, a_2, \cdots , a_N$ are given in $\mathbb{R}^n$ with all lengths at most $1$. Is it possible to put $+$ and $-$ in place of $*$ in the expression $$a_1 * a_2 * \cdots * a_N ,$$ so that the result will have length at most $\sqrt{n}$.