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Following the clarification in the comments, I am interpreting the question as follows.

Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $\ell$ such that the effective Weil divisor $\ell N$ is Cartier and such that the pullback effective Cartier divisor $\pi^*(\ell N)$ equals $\ell A$ as effective Cartier divisors, is $N$ Cartier?

The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.

Let $C\subset \mathbb{P}^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $\mathbb{P}^3$ over $C$. The minimal desingularization of $Y$, $$\pi:X\to Y,$$ is the closure in $Y\times C$ of the graph of the linear projection from $Y$ to $C$. Denote by $\rho$ the projection, $$ \rho:X\to C.$$ This morphism is a $\mathbb{P}^1$-bundle. The morphism $\rho$ maps the exceptional locus $E$ of $\pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $\rho$. The normal sheaf $\mathcal{O}_X(\underline{E})|_E$ is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(-1)|_C$.

Let $H\subset C$ be the restriction to $C$ of a general hyperplane in $\mathbb{P}^2$. Let $D\subset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $\ell>1$ in the Picard group of $C$, i.e., $\ell H - \ell D$ is the divisor of a rational function $f$ on $C$.

Let $M\subset Y$, resp. $N \subset Y$, be the cone over $H$, resp. $D$. Note that $\ell N$ and $\ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $\mathcal{O}_{\mathbb{P}^3}(\ell)|_Y$. In fact, $\ell$ is the smallest positive integer such that $\ell N$ is a Cartier divisor on $Y$.

Denote by $\widetilde{M}\subset X$, resp. by $\widetilde{N}\subset X$, the strict transform under $\pi$ of $M$, resp. of $N$. Note that $\ell \widetilde{M}-\ell\widetilde{N}$ equals the Cartier divisor of $f\circ \rho$, as does the total pullback under $\pi$ of the Cartier divisor $\ell N-\ell M$. Thus, the coefficient of $E$ in the pullback of $\ell M$ equals the coefficient of $E$ in the pullback of $\ell N$.

Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $\widetilde{M}$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $\ell M$ equals $\ell \widetilde{M} + \ell E$.

Therefore also $\ell N$ equals t$\ell \widetilde{N} +\ell E$. Although $N$ is not Cartier on $Y$, the pullback of $\ell N$ equals $\ell A$ on $X$ for the effective Cartier divisor $A=\widetilde{N}+E$.

Following the clarification in the comments, I am interpreting the question as follows.

Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $\ell$ such that the effective Weil divisor $\ell N$ is Cartier and such that the pullback effective Cartier divisor $\pi^*(\ell N)$ equals $\ell A$ as effective Cartier divisors, is $N$ Cartier?

The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.

The minimal resolution of a cone over a smooth plane cubic is a ruled surface. Let $C\subset \mathbb{P}^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $\mathbb{P}^3$ over $C$. The minimal desingularization of $Y$, $$\pi:X\to Y,$$ is the closure in $Y\times C$ of the graph of the linear projection from $Y$ to $C$. Denote by $\rho$ the projection, $$ \rho:X\to C.$$ This morphism is a $\mathbb{P}^1$-bundle. The morphism $\rho$ maps the exceptional locus $E$ of $\pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $\rho$. The normal sheaf $\mathcal{O}_X(\underline{E})|_E$ is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(-1)|_C$.

Effective Weil divisors "torsion-equivalent" to a conical hyperplane class. Let $H\subset C$ be the restriction to $C$ of a general hyperplane in $\mathbb{P}^2$. Let $D\subset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $\ell>1$ in the Picard group of $C$, i.e., $\ell H - \ell D$ is the divisor of a rational function $f$ on $C$.

Let $M\subset Y$, resp. $N \subset Y$, be the cone over $H$, resp. $D$. Note that $\ell N$ and $\ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $\mathcal{O}_{\mathbb{P}^3}(\ell)|_Y$. In fact, $\ell$ is the smallest positive integer such that $\ell N$ is a Cartier divisor on $Y$.

Denote by $\widetilde{M}\subset X$, resp. by $\widetilde{N}\subset X$, the strict transform under $\pi$ of $M$, resp. of $N$. Note that $\ell \widetilde{M}-\ell\widetilde{N}$ equals the Cartier divisor of $f\circ \rho$, as does the total pullback under $\pi$ of the Cartier divisor $\ell N-\ell M$. Thus, the coefficient of $E$ in the pullback of $\ell M$ equals the coefficient of $E$ in the pullback of $\ell N$.

Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $\widetilde{M}$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $\ell M$ equals $\ell \widetilde{M} + \ell E$.

Therefore also $\ell N$ equals t$\ell \widetilde{N} +\ell E$. Although $N$ is not Cartier on $Y$, the pullback of $\ell N$ equals $\ell A$ on $X$ for the effective Cartier divisor $A=\widetilde{N}+E$.

Following the clarification in the comments, I am interpreting the question as follows. For an effective Weil divisor $N$ on $Y$ such that for some positive integer $\ell$ the effective Weil divisor $\ell N$ is Cartier, if the pullback effective Cartier divisor $\pi^*(\ell N)$ equals $\ell A$ for an effective Cartier divisor $A$ on $X$ (equality as integral effective Cartier divisors, not just as linear equivalence classes of Cartier divisors), does it follow that $N$ is Cartier on $Y$? The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.

Let $Y$ be the projective cone in $\mathbb{P}^3$ over a smooth plane cubic curve $C\subset \mathbb{P}^2$. Let $\pi:X\to Y$ be the minimal desingularization. The linear projection from $Y$ to $C$ gives a regular morphism, $$\rho:X\to C,$$ that is a $\mathbb{P}^1$-bundle. The morphism $\rho$ maps the exceptional locus $E$ of $\pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $\rho$. The normal sheaf $\mathcal{O}_X(\underline{E})|_E$ is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(-1)|_C$.

Let $H\subset C$ be the restriction to $C$ of a general hyperplane in $\mathbb{P}^2$. Let $D\subset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $\ell>1$ in the Picard group of $C$, i.e., $\ell H - \ell D$ is the divisor of a rational function $f$ on $C$.

Let $M\subset Y$, resp. $N \subset Y$, be the cone over $H$, resp. $D$. Note that $\ell N$ and $\ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $\mathcal{O}_{\mathbb{P}^3}(\ell)|_Y$. In fact, $\ell$ is the smallest positive integer such that $\ell N$ is a Cartier divisor on $Y$.

Denote by $\widetilde{M}\subset X$, resp. by $\widetilde{N}\subset X$, the strict transform under $\pi$ of $M$, resp. of $N$. Note that $\ell \widetilde{M}-\ell\widetilde{N}$ equals the Cartier divisor of $f\circ \rho$, as does the total pullback under $\pi$ of the Cartier divisor $\ell N-\ell M$. Thus, the coefficient of $E$ in the pullback of $\ell M$ equals the coefficient of $E$ in the pullback of $\ell N$.

Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $\widetilde{M}$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $\ell M$ equals $\ell \widetilde{M} + \ell E$.

Therefore also $\ell N$ equals t$\ell \widetilde{N} +\ell E$. Although $N$ is not Cartier on $Y$, the pullback of $\ell N$ equals $\ell A$ on $X$ for the effective Cartier divisor $A=\widetilde{N}+E$.

Following the clarification in the comments, I am interpreting the question as follows.

Question. For an effective Weil divisor $N$ on $Y$, for an effective Cartier divisor $A$ on $X$, for a positive integer $\ell$ such that the effective Weil divisor $\ell N$ is Cartier and such that the pullback effective Cartier divisor $\pi^*(\ell N)$ equals $\ell A$ as effective Cartier divisors, is $N$ Cartier? 

The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.

Let $C\subset \mathbb{P}^2$ be a smooth, plane cubic (a genus $1$ curve). Let $Y$ be the projective cone in $\mathbb{P}^3$ over $C$. The minimal desingularization of $Y$, $$\pi:X\to Y,$$ is the closure in $Y\times C$ of the graph of the linear projection from $Y$ to $C$. Denote by $\rho$ the projection, $$ \rho:X\to C.$$ This morphism is a $\mathbb{P}^1$-bundle. The morphism $\rho$ maps the exceptional locus $E$ of $\pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $\rho$. The normal sheaf $\mathcal{O}_X(\underline{E})|_E$ is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(-1)|_C$.

Let $H\subset C$ be the restriction to $C$ of a general hyperplane in $\mathbb{P}^2$. Let $D\subset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $\ell>1$ in the Picard group of $C$, i.e., $\ell H - \ell D$ is the divisor of a rational function $f$ on $C$.

Let $M\subset Y$, resp. $N \subset Y$, be the cone over $H$, resp. $D$. Note that $\ell N$ and $\ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $\mathcal{O}_{\mathbb{P}^3}(\ell)|_Y$. In fact, $\ell$ is the smallest positive integer such that $\ell N$ is a Cartier divisor on $Y$.

Denote by $\widetilde{M}\subset X$, resp. by $\widetilde{N}\subset X$, the strict transform under $\pi$ of $M$, resp. of $N$. Note that $\ell \widetilde{M}-\ell\widetilde{N}$ equals the Cartier divisor of $f\circ \rho$, as does the total pullback under $\pi$ of the Cartier divisor $\ell N-\ell M$. Thus, the coefficient of $E$ in the pullback of $\ell M$ equals the coefficient of $E$ in the pullback of $\ell N$.

Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $\widetilde{M}$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $\ell M$ equals $\ell \widetilde{M} + \ell E$.

Therefore also $\ell N$ equals t$\ell \widetilde{N} +\ell E$. Although $N$ is not Cartier on $Y$, the pullback of $\ell N$ equals $\ell A$ on $X$ for the effective Cartier divisor $A=\widetilde{N}+E$.

Following the clarification in the comments, I am interpreting the question as follows. For an effective Weil divisor $N$ on $Y$ such that for some positive integer $\ell$ the effective Weil divisor $\ell N$ is Cartier, if the pullback effective Cartier divisor $\pi^*(\ell N)$ equals $\ell A$ for an effective Cartier divisor $A$ on $X$ (equality as integral effective Cartier divisors, not just as linear equivalence classes of Cartier divisors), does it follow that $N$ is Cartier on $Y$? The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.

Let $Y$ be the projective cone in $\mathbb{P}^3$ over a smooth plane cubic curve $C\subset \mathbb{P}^2$. Let $\pi:X\to Y$ be the minimal desingularization. The linear projection from $Y$ to $C$ gives a regular morphism, $$\rho:X\to C,$$ that is a $\mathbb{P}^1$-bundle. The morphism $\rho$ maps the exceptional locus $E$ of $\pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $\rho$. The normal sheaf $\mathcal{O}_X(\underline{E})|_E$ is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(-1)|_C$.

Let $H\subset C$ be the restriction to $C$ of a hyperplane. Let $D\subset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $\ell>1$ in the Picard group of $C$, i.e., $\ell H - \ell D$ is the divisor of a rational function $f$ on $C$.

Let $M\subset Y$, resp. $N \subset Y$, be the cone over $H$, resp. $D$. Denote by $\widetilde{M}\subset X$, resp. by $\widetilde{N}\subset X$, the strict transform under $\pi$ of $M$, resp. of $N$. Note that $\ell \widetilde{M}-\ell\widetilde{N}$ equals the divisor of $f\circ \rho$, as does the total pullback under $\pi$ of the Cartier divisor $\ell N-\ell M$. Thus, the coefficient of $E$ in the pullback of $M$ equals the coefficient of $E$ in the pullback of $N$.

The pullback of $M$ is the strict transform $\widetilde{M}$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$.

Therefore also $\ell N$ equals the sum of $\ell \widetilde{N}$ plus $\ell E$. Although $N$ is not Cartier on $Y$, the pullback of $\ell N$ equals $\ell A$ on $X$ for the effective Cartier divisor $A=\widetilde{N}+E$.

Following the clarification in the comments, I am interpreting the question as follows. For an effective Weil divisor $N$ on $Y$ such that for some positive integer $\ell$ the effective Weil divisor $\ell N$ is Cartier, if the pullback effective Cartier divisor $\pi^*(\ell N)$ equals $\ell A$ for an effective Cartier divisor $A$ on $X$ (equality as integral effective Cartier divisors, not just as linear equivalence classes of Cartier divisors), does it follow that $N$ is Cartier on $Y$? The answer to that question is no. The following example is a modification of the example in my comment avoiding the mistake identified by Stefano.

Let $Y$ be the projective cone in $\mathbb{P}^3$ over a smooth plane cubic curve $C\subset \mathbb{P}^2$. Let $\pi:X\to Y$ be the minimal desingularization. The linear projection from $Y$ to $C$ gives a regular morphism, $$\rho:X\to C,$$ that is a $\mathbb{P}^1$-bundle. The morphism $\rho$ maps the exceptional locus $E$ of $\pi$ isomorphically to $C$, and $E$ is a relative hyperplane class for $\rho$. The normal sheaf $\mathcal{O}_X(\underline{E})|_E$ is isomorphic to $\mathcal{O}_{\mathbb{P}^2}(-1)|_C$.

Let $H\subset C$ be the restriction to $C$ of a general hyperplane in $\mathbb{P}^2$. Let $D\subset C$ be a degree $3$ effective divisor such that the divisor $H-D$ has finite order $\ell>1$ in the Picard group of $C$, i.e., $\ell H - \ell D$ is the divisor of a rational function $f$ on $C$.

Let $M\subset Y$, resp. $N \subset Y$, be the cone over $H$, resp. $D$. Note that $\ell N$ and $\ell M$ are linearly equivalent Cartier divisors on $Y$ in the linear equivalence class of $\mathcal{O}_{\mathbb{P}^3}(\ell)|_Y$. In fact, $\ell$ is the smallest positive integer such that $\ell N$ is a Cartier divisor on $Y$.

Denote by $\widetilde{M}\subset X$, resp. by $\widetilde{N}\subset X$, the strict transform under $\pi$ of $M$, resp. of $N$. Note that $\ell \widetilde{M}-\ell\widetilde{N}$ equals the Cartier divisor of $f\circ \rho$, as does the total pullback under $\pi$ of the Cartier divisor $\ell N-\ell M$. Thus, the coefficient of $E$ in the pullback of $\ell M$ equals the coefficient of $E$ in the pullback of $\ell N$.

Already $M$ is an effective Cartier divisor on $Y$, and the pullback of $M$ is the strict transform $\widetilde{M}$ plus the exceptional divisor $E$. One way to see this is to deform $M$ to a hyperplane section of $Y$ that is disjoint from the vertex of the cone. Since the intersection number of the total transform of $M$ with $E$ equals $0$, it follows that the coefficient of $E$ equals $1$. Thus, the total pullback of $\ell M$ equals $\ell \widetilde{M} + \ell E$.

Therefore also $\ell N$ equals t$\ell \widetilde{N} +\ell E$. Although $N$ is not Cartier on $Y$, the pullback of $\ell N$ equals $\ell A$ on $X$ for the effective Cartier divisor $A=\widetilde{N}+E$.