You pick $N-1$ points $X_1,\dotsc, X_{N-1}$ independently and uniformly at random in $[0,1]$. They divide the segment $[0,1]$ into $N$ segments of lengths $S_1,\dotsc, S_N$ usually referred to as *spacings*.

To use your description, rearrange the $X_i$ in increasing order $X_{(1)}\leq \cdots \leq X_{(N-1)}$ and then

$$ S_1= X_{(1)},\;\;S_2= X_{(2)}-X_{(1)},\dotsc , S_N=1-X_{(N-1)}. $$

The random vector $\vec{S}=(S_1,\dotsc, S_n)$ is uniformly distributed on the $(N-1)$-dimensional simplex $\newcommand{\bR}{\mathbb{R}}$

$$ \Delta_{N-1}= \bigl\{\; (s_1,\dotsc, s_N)\in \bR^N_+;\;\;\sum_{j=1}^N s_j=1\;\bigr\} $$

equipped with the normalized surface measure. The "area" $A_{N-1}$ of $\Delta_{N-1}$ is found using the recurrence

$$A_1=\sqrt{2},\;\;A_N= \sqrt{\frac{N+1}{N}}\frac{A_{N-1}}{N}. $$

This yields

$$A_N= \frac{\sqrt{N+1}}{N!}. $$

Rearrange the spacings $S_1,\dotsc,, S_n$ in increasing order

$$S_{(1)}\leq \cdots \leq S_{(N)}. $$

The random vector $(S_{(1)},\dotsc, S_{(N)})$ is uniformly distributed in the $N-1$-dimensional polyhedron

$$ P_{N-1}=\bigl\{\, (s_1,\dotsc, s_N)\in \Delta_{N_1};\;\;s_1\leq \cdots \leq s_N\;\bigr\}. $$

This is another $(N-1)$-dimensional simplex with vertices

$$v_1=(0,0,\dotsc, 0,1),\;\;v_2=\frac{1}{2}(0,\dotsc, 0,1,1),\;v_3=\frac{1}{3}(0,\dotsc, 0,1,1,1),\dotsc, $$

and total "area"

$$ \alpha_{N-1}=\frac{1}{N!} A_{N-1}. $$

The mean you are looking for is the center of mass of the simplex $P_{N-1}$, where the mass is uniformly distributed on this polyhedron.

To compute effectively the coordinates of this center of mass it may be convenient to work with new linear coordinates, $x=(x_1,\dotsc, x_N)$ determined by the basis $v_1,\dotsc, v_N$ of $\bR^N$. They are related to the standard Euclidean coordinates $(s_1,\dotsc, s_N)$ of $\bR^N$ via the equalities

$$ s_j=s_j(x) =\sum_{k=N-j+1}^N\frac{1}{k}x_k,\;\;j=1,\dotsc, N.$$

In the $x$-coordinates the simplex $P_{N-1}$ becomes the simplex

$$ D_{N-1}= \bigl\{\; (x_1,\dotsc, x_N)\in \bR^N_+;\;\;\sum_{j=1}^N x_j=1\;\bigr\} $$

Denote by $dA_x$ the *normalized* area form on $D_{N-1}$. The coordinates of the center of mass of $P_{N-1}$ are $(\bar{s}_1,\dotsc, \bar{s}_N)$, where

$$\bar{s}_j=\int_{D_{N-1}} s_j(x) dA_x,\;\;j=1,\dotsc, N. $$

Now observe that

$$\int_{D_{N-1}} x_j dA_x= \frac{1}{N},\;\;j=1,\dotsc, N.$$

We deduce

$$ \boxed{E[S_{(j)}]=\bar{s}_j=\frac{1}{N}\sum_{k=N-j+1}^N\frac{1}{k},\;\;j=1,\dotsc, N.} $$

Here is also a plausibility test. The above formula implies

$$\bar{s}_1+\cdots + \bar{s}_N=1. $$

This is in perfect agreement with the equality $S_{(1)}+\cdots +S_{(N)}=1$.

**Remark.** The problem of the distribution of spacings is rather old. For example there is *Witworth formula* (1897) describing the expectation of the larges spacing $S_{N}$. More precisely it states

$$ E[S_{(N)}]=\underbrace{\sum_{k=1}^{N}(-1)^{k+1} \frac{1}{k^2} \binom{N-1}{k-1}}_{=:L_N}. $$

It looks rather different from what we proved

$$ E[S_{(N)}]=\bar{s}_N=\frac{1}{N}\left(1+\frac{1}{2}+\cdots +\frac{1}{N}\right). $$

I have not thought how prove directly that

$$ L_N=\bar{s}_N, $$

but simple computer simulations show that indeed the two rather different descriptions are identical. For more details about the spacings problem see this blogpost.