$X$, $Y$ - independent random variables such that $$P(X=x)=\dfrac{1}{n}=P(Y=x)$$ for $x \in \{1,\ldots, n\}$. Calculate $\mathbb{E}[\min(X,Y)\mid Y].$
\begin{align} \psi(j)&=\sum_i i\cdot P(\min(X,Y)=i\mid Y=j) \\&=\dfrac{\sum_i i\cdot P(\min(X,Y)=i \land Y=j)}{P(Y=j)} \\&=\dfrac{\sum_i i\cdot P(\min(X,Y)=i \land Y=j)}{\frac{1}{n}} \\&=n\cdot \sum_i i\cdot P(\min(X,Y)=i \land Y=j) \\&=n\cdot\left(\sum_{i\leq j} i\cdot P(\min(X,Y)=i \land Y=j) + \sum_{i> j} i\cdot P(\min(X,Y)=i \land Y=j)\right) \end{align}
Do $\sum_{i\leq j} i\cdot P(\min(X,Y)=i \land Y=j) = \sum_{i\leq j} i\cdot P(X=i \land Y=j)$?
How to compute this sum. What to do with 2nd sum?
If $i<j$, then $$P(\min(X,Y)=i\: , Y=j) = P(X=i\:,Y=j).$$ If $i=j$, then $$P(\min(X,Y)=i \:, Y=j) = P(X\geq j \:, Y=j).$$ For the last sum you can notice, that $\min(X,Y) \leq Y$ with probability $1$, so for $i>j$ we have $$P(\min(X,Y)=i\: , Y=j) = P(\emptyset)=0.$$ Plugging this into your established formula, you get \begin{align*} \psi(j) &= n\Big(\sum_{i=1}^{j-1} iP(X=i,Y=j)+jP(X\geq j,Y=j)\Big) \\ &= n\Big(\sum_{i=1}^{j-1} i \frac{1}{n^2}+\frac{j(n-j+1)}{n^2}\Big) \\ &= \frac{(j-1)j}{2n}+\frac{j(n-j+1)}{n} \end{align*} I however would prefer the following method: \begin{align*} E[\min(X,Y) \: | \: Y=j] &= E[\min(X,j) \: | \: Y=j] \\ &= E[\min(X,j)] \quad \quad \quad \quad \text{(by independence)} \\ &= \frac{1}{n}\sum_{i=1}^n min(i,j) \\ &= \frac{1}{n} \Big(\sum_{i=1}^{j-1} i + \sum_{i=j}^n j \Big) \\ &= \frac{(j-1)j}{2n}+\frac{j(n-j+1)}{n} \end{align*}