Given x was randomly picked from $X={1,2, \cdots n}$ and that Y was randomly picked such that $Y<X$. I am interested in expected value of Y.
From my calculations:
$$ P(Y=j ) = \sum_{k} P(Y=j \mid k) P(X=k) = \frac{1}{n} \sum_{k=1}^n \frac{1}{k} $$
$$ P(Y=j \mid X=j ) = \frac{P(X=j,Y=j)}{P(X=j)}= \frac{1}{j} $$
For the expectation, I could do: $$ E(Y) = \sum_{j=1}^n y_j P(Y=y_j) = \frac{1}{n} \sum_{j=1}^n y_j \sum_{k=1}^n \frac{1}{k} $$
And also do: $$E(Y\mid X) = \sum_{k=1}^n y_k P(Y\mid X) = \sum_{k=1}^n y_k \frac{1}{j} $$ $$ E(E(Y\mid X)) = \sum_{j=1}^n x_j E(Y\mid X) = \sum_{j=1}^n x_j \sum_{k=1}^n y_k \frac{1}{j} $$
I think simpler approach is:
$$E(Y\mid X) = \frac{(X+1)}{2} $$ Since given $X$, $Y$ is uniform from $\{1,2 \cdots X\}$
$$ E(E(Y\mid X)) =E(\frac{X+1}{2}) = \frac{1}{2}\left(E(X) +E(1) \right) = \frac{1}{2} \left (\frac{n+1}{2} +1\right)= \frac{n+3}{4} $$
Does this makes sense?
I would like to establish a correspondence between this discrete question and its natural continuous "counterpart".
Consider that $X \approx Unif([0,1])$.
Saying that $Y$ is chosen in a uniform way in $[0,X]$ is equivalent to writing it as the product of 2 independent uniform random variables $X,Z \approx Unif([0,1]):$
$$Y=ZX$$
Consequently, due to independence:
$$\mathbb{E}(Y)=\mathbb{E}(Z)\mathbb{E}(X)=(\tfrac12)^2=\tfrac14$$
Now, if instead of a $[0,1]$ range, one has a $[1,n]$ range, a natural discrete-to-continuous analogy gives, by proportionality, a mathematical expectation equivalent to $\frac{n}{4}$ (for large values of $n$) which is indeed the case for your result $\frac{n+3}{4}.$
Remark : This continuous reasoning could hopefuly be adapted into a proof of this question remaining in the discrete world...