Well, you did not define and detail well the problem, hence, I will first write formally the problem definition based on my understanding of what you have written, and then I will propose an Integer Programming formulation.
Problem definition
Let's first define formally the problem.
Let:
- $t$ be the number of tasks to be serviced;
- $T = \mathbb{N}_{\leqslant t}^{*}$ be the set of tasks to be serviced;
- $m$ be the number of available weeks;
- $M = \mathbb{N}_{\leqslant m}^{*}$ be the set of available weeks;
- $w_i \in \mathbb{Z}$ be the number of weeks the task $i \in T$ requires;
- $p_{il} \in \mathbb{Z}$ be the number of workers the task $i \in T$ requires in week $l \in M$.
A feasible solution for this variant of the classical scheduling problem is given by an allocation of the tasks in $T$ in the weeks of $M$ in a way that, if the task $i \in T$ starts to be executed in the week $l \in M: l \leqslant m - w_i + 1$, then the task $i$ will be allocated, in order to guarantee the continuity of the task execution, on the weeks in {$l^{'} \in M: l \leqslant l^{'} < l + w_i$}. An optimal solution for this variant is a feasible solution in which the maximum amount of workers used in any week is minimum.
An Integer Programming formulation
Let's consider the variables
$$
x_{il} = \begin{cases}
1 &\text{if task $i \in T$ is executed in week $l \in M$,}\\
0 &\text{otherwise}
\end{cases},
\\
z_{il} = \begin{cases}
1 &\text{if task $i \in T$ starts to be serviced in week $l \in M : l \leqslant m - w_i + 1$,}\\
0 &\text{otherwise}
\end{cases},
\\
\text{and}
\\
y \in \mathbb{Z} \text{ is the number of workers that the week with the}
\\
\text{maximum number of workers has in a feasible solution}.
$$
Below follows the IP formulation.
Objective Function:
$$
\text{min } y
$$
Constraints:
The variable $y$ assumes a value greater than or equals to the number of workers used in each week.
$$
y \geqslant \sum_{i \in T} p_{il} x_{il} \quad \forall l \in M;
$$
If the task $i \in T$ begins to be serviced in the week $l \in M : l \leqslant m - w_i + 1$, then this task will be allocated to all the weeks in $\{l^{'} \in M : l \leqslant l^{'} < l + w_i\}$.
$$
z_{il} = x_{il^{'}} \quad \forall i \in T,
\\
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \forall l \in M : l \leqslant m - w_i + 1,
\\
\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \forall l^{'} \in M : l \leqslant l^{'} < l + w_i;
$$
A task $i \in T$ must start to be serviced in exactly one feasible week, i.e., one week between $1$ and $m - w_i + 1$.
$$
\sum_{l \in M : l \leqslant m - w_i + 1} z_{il} = 1 \quad \forall i \in T.
$$
If you want to consider a serial, non-parallel, version of the problem. You can add the following set of constraints to the formulation.
$$
\sum_{i \in T} x_{il} \leqslant 1 \quad \forall l \in M.
$$
If anyone has any question or have found any error, please let me know.
