The answer is $$ f(n) = \begin{cases} 3t^2 - t & \text{ if } x = 3t-1 \\ 3t^2 + t & \text{ if } x = 3t \\ 3t^2 + 3t + 1 & \text{ if } x = 3t+1 \\ \end{cases}$$ Bound. Let $ k = \lfloor \frac n3 \rfloor$. We only consider (the union of) bars of size $1 \times (k+1), \cdots, 1 \times n$. Some are vertical, and some are horizontal. Note if two bars have the same orientation, then they are disjoint. Thus, we have \begin{align*}\#\text{cells} &\geq (k+1)+(k+2)+\dots+n - \#(\text{overlap cells}) \\ &= \frac{(n+k+1)(n-k)}2 - \#(\text{horizontal bars}) \cdot \#(\text{vertical bars}) \\ &\geq \frac{(n+k+1)(n-k)}2 - \left\lfloor \frac{(n-k)^2}4\right\rfloor \end{align*}We can check with casework with $n$ mod 3 that this shows the number of cells is at least $f(n)$; note $n-k$ is odd iff $n\equiv 1\pmod 3$. Construction The construction basically follows the equality case above. Here is the diagram for $n=21$. [asy][asy] size(5.5cm); int n = 21; int k = n # 3; int l = (n-k) # 2; for(int i=0; i<=n; ++i){ draw((i,0)--(i,-(k+l)), mediumgray); } for(int j=0; j<=k+l; ++j){ draw((0,-j)--(n,-j), mediumgray); } for(int i=0; i<n-k-l; ++i){ draw(box((0,-i),(n-i, -(i+1))), linewidth(1)); } for(int i=0; i<l; ++i){ draw(box((i,0), (i+1,-(k+l-i))), linewidth(1)); } [/asy][/asy] Let $k=\lfloor\tfrac n3\rfloor$, and let $l = \lfloor\tfrac{n-k}2\rfloor$. We then Stack bars of size $1\times n$, $\dots$, $1\times (k+l+2)$, $1\times (k+l+1)$ horizontally (from top to bottom) so that their leftmost cells share the same column. Stack bars of size $(k+1)\times 1$, $(k+2)\times 1$, $\dots$, $(k+l)\times 1$ vertically so that their topmost cells is exactly the top bar. Bars of size $1\times 2$, $1\times 3$, $\dots$, $1\times k$ are guaranteed to be included. This construction matches all equalities in the bound above.

The answer is $3k^2 + 3k + 1$ for $3k + 1$, $3k^2 + 5k + 2$ for $3k + 2$, and $3k^2 + 7k + 4$ for $3k + 3$. Construction Take the base case of $k = 0$ by taking a $1 \times 1, 1 \times 2$ and an $L$ shape. We strengthen to the claim of also containing a $1 \times 1$ bar. To induct up from $t$ to $t + 3$, attach an $L$ shape with legs of length $t+3, t+2$ (for a total of $t+2$ blocks) to the Young Tableaux. This increases the length of all the earlier bars by $1$, adds a new $1 \times 1$ bar, and adds a $1 \times t+3$ and $1 \times t+2$ length bar. [asy][asy] size(5cm); pen pa, pb; pa = purple+white; pb = green+white; filldraw(unitsquare, pa); filldraw(shift(1,0)*unitsquare, pa); filldraw(shift(2,0)*unitsquare, pa); filldraw(shift(3,0)*unitsquare, pa); filldraw(shift(4,0)*unitsquare, pa); filldraw(shift(0,1)*unitsquare, pa); filldraw(shift(0,2)*unitsquare, pa); filldraw(shift(0,3)*unitsquare, pa); filldraw(shift(0,4)*unitsquare, pa); filldraw(shift(0,5)*unitsquare, pa); filldraw(shift(1,1)*unitsquare, pb); filldraw(shift(1,2)*unitsquare, pb); filldraw(shift(1,3)*unitsquare, pb); filldraw(shift(2,1)*unitsquare, pb); [/asy][/asy] Bounding Suppose that bars of length $1 \times 2, \dots, 1 \times t$ are required. For each bar, mark some row or column that contains it. Let $r$ be the number of rows and $c$ be the number of columns. WLOG let $r < c$. Now, consider the $r-2$ bars with length less than $r$. These bars can be only filled to their max. If $r' < r, c' < c$ are the rows which contain a bar that isn't one of the $r-2$, then there are at least \[ \frac{t(t+1)}{2} - 1 - r'c' - (2 + \dots + r-1) \]elements. Note that $r' + c' \le (t-r+1)$ so $r'c' \le \min\{\frac{(t-r+1)^2}{4}, r(t+1-2r) \}$. This gives us one bound \[ \frac{t(t+1)}{2} - 1 - \frac{(t-r+1)^2 + 2((r-1)r - 2)}{4} \]which is concave in $r$, so taking the smallest $r$ possible works. On the other hand, \[ \frac{t(t+1)}{2} - 1 - \frac{4r(t+1-2r) + 2((r-1)r - 2)}{4} \]is minimized at $r = \frac{2t+1}{6}$ As such, the function is minimized at the point we swap between the two $\min$ functions, or $r = \frac{t}{3} = \frac{t+1}{3}$. We then get a value of at least \begin{align*} \frac{t(t+1)}{2} - 1 - \frac{t+1}{3} \cdot \left(t+1 - \frac{2(t+1)}{3}\right) - \frac{(t+1)(t-2)}{18} + 1 &= \\ \frac{t(t+1)}{2} - \frac{(t+1)^2}{9} - \frac{(t+1)(t-2)}{18} &= \\ \frac{1}{18} \left(9t(t+1) - 2(t+1)^2 - (t+1)(t-2)\right) &= \\ \frac{1}{18} \left(9t^2 + 9t - 2t^2 - 4t - 2 - t^2 + t + 2\right) &= \\ \frac{t^2}{3} + \frac{t}{3} \\ \end{align*}Then we have that $\frac{(3k+1)^2}{3} + \frac{3k+1}{3} = 3k^2 + 3k + \frac{2}{3}$, $\frac{(3k+2)^2}{3} + \frac{3k+2}{3} = 3k^2 + 5k + 2$, and $\frac{(3k+3)^2}{3} + \frac{3k+3}{3} = 3k^2 + 7k + 4$. Remark: This is actually why equality holds for the latter two cases but not for the first.